Các dạng bài tập tích phân

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Nội dung Text: Các dạng bài tập tích phân

Chapter 5e Integral of Irrational Function

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dx
8/ I = −
( x − q) ax 2 + bx + c
dx 1 dt 1
= Put x − q = � dx = − , t =
x−q
t2
t
( x − q) ax 2 + bx + c
2
� ��
1 1 1 2q 1
� �� � �
2
+ q2 � b � + q � c
ax + bx + c = a � + q � + b � + q � c = a � +
+ + +
2
t t t t
� �� � � �
� �
t

))
( (
b + 2aq
a 1
a + t ( b + 2aq ) + t 2 aq 2 + bq + c
+ aq 2 + bq + c =
= +
t2 t2
t

dt

t2
dx
=�
��
))
( (
( x − q ) ax 2 + bx + c 1 1 a + t ( b + 2aq ) + t 2 aq 2 + bq + c
t t2
dt

t2 dt
=� = −�
)) )
( ((
( )
1
a + t ( b + 2aq ) + t 2 aq 2 + bq + c t 2 aq 2 + bq + c + t ( b + 2aq ) + a
2
t




2
( )
dt
Put aq 2 + bq + c = A,
= −+
� �
�2 + t ( b + 2aq ) +
1 a �
t
( ) ( )( )
� �
aq 2 + bq + c � aq 2 + bq + c aq 2 + bq + c �
� �
dt
= −+
2
� 2�
1 � ( b + 2aq ) � a ( b + 2aq )


�+ �+ −
t
4A 2
A� �
2A �A

� �
2 2
� ( b + 2aq ) � a ( b + 2q ) 4Aa − ( b + 2aq )
Put � + � y � dy = dt, N = −
= =
t
( )
4A 2 2
2A � A
� 4 aq 2 + bq + c

dy Ady 2
(b 2aq ) ,
n2
I ++= ��− = =
�2− � If N 0 N 4Aa
( ) ( y2 + N )
y +N
A

)
) ( b2
(
2 2
( b 2aq ) 4abq 4 ( aq )
4Aa �+ �++ �++ ۳ 4a a.q 2 b2
b.q c 4ac
2

� b� c b �
2
2 2
A > 0 � A = m � a.q + b.q + c � � a �q + � + − � 0�
0 �
� 2a � a 4 �

� �
2
�� b � 4ac − b �
2
�q + � + � a>0
�a � �
2a � 4

� �
� �


( y2 + n 2 ) �
m.dy �
� I = −+ = −m.ln � +
y �
( ) � �
2 2
y +n

b + 2q
� B� 1
y = �+ = +
t
( )

� 2A � x − q 2 aq 2 + bq + c




3
2� �
2
� ( b + 2aq ) � a ( b + 2aq ) ( b + 2aq ) a
= �2 + t �
y 2 + n 2 = �+ �+ − +
t t
( )( )
� �
4A 2 aq 2 + bq + c aq 2 + bq + c
2A � A
� � �
� �

)
( t ( aq )
+ bq + c + t ( b + 2aq ) + a
2 2
=
( aq2 + bq + c)
)�
(
�( )
�t 2 aq 2 + bq + c + t ( b + 2aq ) + a �
t2
=
( aq2 + bq + c) � �
t2
� �
� �


)
( t ( aq )
+ bq + c + t ( b + 2aq ) + a
2 2
*
t2
2
� 2� �
1 2q 1 1 1
� � �� �
= a� + + q � b� + q � c = a � + q�+ b� + q� c
+ + +
�2 t t t t
�� � � �� �
t
ax 2 + bx + c
= ax 2 + bx + c � y 2 + n 2 =
( x − q ) 2 ( aq 2 + bq + c )
�1 1�
x = +q�t =
� �
x−q�
�t


( y2 + n 2 ) �

ln � +
y �
dx � �
�I=� =−
1
( x − q ) ax 2 + bx + c
aq 2 + bq + c
� �
� �
ax 2 + bx + c � 2
b + 2aq
�1 1
� �
= − ln � + + � aq + bq + c
( ) ( )
� ( x − q)

� x − q 2 aq + bq + c
2
aq 2 + bq + c �
� �
� �
� �




4
dx
* I=−
( x − 1) 1 − x2
dx 1
* I=� Put : x − 1 =
t
( x − 1) 1 − x 2
2
−dt 1 + 2t
� 1�
2
� dx = 1 − x = 1 − �+ � = −
1
t2 t2
� t�

dt
2 dt
� I = −� t =�
−1 − 2t −1 − 2t
t. t

( because 1 − x 2 > 0 � x < 1 � x − 1 < 0 � t < 0 � t = −t )
1
− +1
1 d ( −2t − 1) 1 ( −1 − 2t ) 2 1− x
2
I=− =− = − −1 − 2t = − −1 − =− +C
2 � −1 − 2t 1 x −1 1+ x
2 − +1
2

3x + 4 2
− x 2 + 6x − 8 = 1 − ( x − 3)
VD : = dx
− x 2 + 6x − 8
Put x − 3 = t � x = t + 3 � dx = dt
3( 3 + t ) + 4
3x + 4
dx = �
�2 dx
2
− x + 6x − 8 1− t

( )
2
3 d 1− t
3t.dt dt
=� + 13� =− � + 13arcsin t
2
2 2 2
1− t 1− t 1− t

( d ( 1 − t ) = −2t.dt ) = −3 1 − t
2 2
+ 13arcsin t

= −3 − x 2 + 6x − 8 + 13arcsin ( x − 3) + C

R� �
2
For evaluating integral �� ax + bx + c � e can make a trigonometric change of
x, w
� �
variables:




5
2
�� b� c b �
2
2
ax + bx + c = a �x + � + − �

�� 2a � a 4a 2 �
� �
2

� b � b − 4ac �
2
( )
� a u 2 − d 2 If b 2 − 4ac > 0, a > 0
= a �x + � − =

� 2�
� 2a � 4a
� �
2

� b � 4ac − b �
2
( )
� a u 2 + d 2 If b2 − 4ac < 0, a > 0
=a ��x + � + =
� 4a 2 �
� 2a �
� �
2
� 2 − 4ac � b ��
( )
b
− � + � � −a d 2 − u 2 If b 2 − 4ac > 0, a < 0
= −a � =
x
� 4a 2 � 2a ��
� �

d
�ax 2 + bx + c � �� u 2 − d 2 � Put u =
=
��
R x, � R�u, �
sin t
� � � �
R� � R� 2�
2 2
�� ax + bx + c � �� d − u � Put u = d.sin t
=
x, u,
� � � �
R� � R� 2�
2 2
�� ax + bx + c � �� u + d � Put u = d.tgt
=
x, u,
� � � �


( x + 2) 3 �
1�
dx
I=+ = sin � +
arctan �C
3� �
*
( )
3
3
� �
x 2 + 4x + 7

dx du
2
x 2 + 4x + 7 = ( x + 2 ) + 3 Put u = x + 2 � I = �
*I=�
( ) ( )
3 3
x 2 + 4x + 7 u2 + 3


( u 2 + 3) = 3( tan 2 t + 1) = cos3t
3
Put u = 3.tan t � du = ,
cos 2 t
( x + 2) 3 �
3.cos3 t.dt u 3� 1 �
sin t 1 �
�I=� = = sin � =
arctan � sin �
arctan �
3 �3 � �
3 3� 3
2 3
� �
cos t. 3




6
� �
� b ��

� + ��
x
( )

dx 4a � 2a � with a > 0, 4ac −c 2
*I=� = sin �
arctan b 0
( )

( )
2
3 2
a. 4ac − b 4ac − b �

ax 2 + bx + c
� �
a2
� �
2
� 4ac 2 �
( a > 0, 4ac −tb ) � b � � −b �
dx 2 2
*I=� ax + bx + a = a �x + � + � �

0 �
2a � � 4a 2 �
( ax 2 + bx + c ) �
3 � �
� �
� �

� − b2 �
b 4ac 1 du
2
Put u = x + , m = � � I= �
� ,
� 4a 2 �
( u2 + m )
2a a. a 23
� �


( u 2 + m2 ) = m2 ( tan 2 t + 1) = cos t
m.dt m
Put u = m.tan t � du = ,
cos 2 t

cos3 t.dt
1 du 1 m.dt 1 1 sin t
�I= = = =
� � � .
3
a.m 2 cos 2 t. a. a m 2
(u )
a. a a. a
23 a.
2 �m �
2
+m cos t. � �
� t�
cos
� �
� b ��

� + ��
x

1 u� 4a
� � 2a �
= =
sin �
arctan � sin �
arctan
( ) �
� − b2 � � m � a. 4ac − b 2 2
4ac 4ac − b �

a. a. � � � �
� 4a 2 �
a2
� �
� �

c
� � b ��
� �
� + ��
x
+. � � �
dx 4a � 2a ��

� I1 = = lim � �
� .sin arctan
( )
( )
3 c+ +4 � a. 4ac − b 2 m �
� �
b
ax 2 + bx + c � �b
− � �
� �−
� �
2a
2a
� �
� b�
�+ �
c
� �
4a 4a
� 2a � =
= . lim sin � �
arctan
( ) ( )
a. 4ac − b 2 c− +4 � a. 4ac − b 2
m

� �
� �




7
2
� 4ac 2 �
( a > 0, 4ac −tb ) � b � � −b �
dx 2 2
*I=+ ax + bx + a = a �x + � + � �

0 �
� 2a � � a 2 �
( ax ) �
n �
� �
� �
2
+ bx + c

� − b2 �
b 4ac 1 du
, m2 = � +
Put u = x + � I=
� ,
� a2 �
( u2 + m )
2a an 2n
� �


( u 2 + m2 ) = m2 ( tan 2 t + 1) = cos t
m.dt m
Put u = m.tan t � du = ,
cos 2 t
( cos t ) n −2 .dt
1 du 1 m.dt 1
+
�I= = =
� � m n −1
n
(u )
n 2n n
an
�m �
a a
2
cos 2 t. �
+m �
� t�
cos

( cos t ) n −2 .dt
( a > 0, 4ac −,b2 0)
dx 1
*I=� = � m n −1
( ax )
n
an
2
+ bx + c

� − b2 �
4ac b u
m2 = � �u = x + , t = arctan
,
� a2 � 2a m
� �

� �
� �
� x+ b
� �

� �
� �
dx 1 dt t 1 2a
n = 2: I = � = �= = arctan �
( )
� �

ax 2 + bx + c a m a.m 4ac − b2 � � 4ac − b 2 �

a. � �
� �
4a 2 � 2
� 4a �

� �
� 2ax + b �
2
� �
= arctan � �
� �
4ac − b 2 � �
2
� 4ac − b �
� �

1 + cos 2t
dx 1 1
( cos t ) 2 .dt = 2 3 �
n = 4: I = � =
2 3�
.dt
( )
2 2
a .m a .m
ax 2 + bx + c

1� � ��
� u�
1 � sin 2t � � �1
u
= �+ = +
t arctan � � .sin �2.arctan � ��
� 2 3�
m�
2a 2 .m3 � 2 � 2a .m � � �2
m � ��
� �
b
x = − � u = 0, x = +�� u = +�
2a


8
+a
+a
� � ��
� u�
dx 1
� �1
u
�− = +
.�
arctan � � .sin �2.arctan � ��
m �0
( )
2
2a 2 .m3 � � �2
m � ��
� �
2
− b/2a ax + bx + c

� sin π �
π � sin π �
π
1 1
= .� + = .� +
� �
( )
3/2 2 2 3/2 �
2� 2 2�
� − b2 � � 4ac − b
4ac
2a 2 . � � 2
2a .
� 4a 2 �
( 2a ) 3
� �

� sin π �
π
4a
= .� + �
( )
2 3/2 �2 2�
4ac − b


� �
� �
c.n 3
n b
+ = 1 − sin � �
arctan � �
* lim
)
( � �
3/2 � �
n a +. i =1 2
2 � 4ac − b
a ( i.c ) + b.n.i.c + n c
2 2 �
� �
c= +.

1 c i.c
, x i � 0, c] , ∆x =
[
Put f ( x i ) = , xi =
( )
n n
3
ax i 2 + bx i + c

c.n 3
n cn 1
� �
� I2 = lim = lim
)
(
3/2
n +c n i =1 �
n n +c i =1 3
i 2 2�
a ( i.c ) + b.n.i.c + n c
2
a ( i.c ) + b.n.i.c + n 2 c
c= +c c +c
i � �
2
� �
n
� �
c
cn n
1 dx
= lim � ( x i ) .∆x = lim �

= lim f
( )
3/2 n +c 3/2
n = +c n i =1 � c2 +a
i 2 0 ax 2 + bx + c
� i =1
i.c i.c
�� ��
c= +c c +=
i �� � + b � � c � +
a
�� � � � �
n n
� �
� �
� �
b
= 1 − sin � �
arctan � �
� 2�
� �
� 4ac − b �
2
� �


( )
p
m n
+x a + bx dx with m, n, p is rational number. The Russian mathematicant Trebushep
prove that the upper integral only can be expressed in elementary function in 3 follow cases:
1/ p is an interger, when that, put x = t s with s is the least common multiple of m, n.
m +1
is an interger, put a + bx s = t s with s is the denominator of p.
2/
n
m +1
+ p is an interger, put ax − n + b = t s with s is the denominator of p.
3/
n
9
−10
1 �1
� −
2 �4

dx
Ex : I = � = � � + 1�
x x
( )
10
x 4 x +1 � �
� �
So p = −10 is a interger, we have case 1/
Put x = t 4 � dx = 4t 3 .dt
� d ( t + 1) d ( t + 1) �
( t + 1) − 1
4t 3 .dt
dt = 4 � �
�I=� = 4� −
� ( t + 1) 9 � + 1) 10 �

10 10
t 2 ( t + 1) ( t + 1) (t
� �
−4 −1
4 4
= + = +
( ) ( )
8 9 8 9
8 ( t + 1) 9 ( t + 1) 2 4 x +1 9 4 x +1


( )
p
* I = + m . a + b.x n
x .dx where p is a interger

( )
p a −1
Put x = t a � dx = a.t a −1.dt � I = a + m.a . a + b.t n.a
t .t .dt

� m +1 −1 �
) . � Ci b.t n.a i .a p−i �.dt = a � Ci t n.a.i +a ( m +1) −1.bi a p −i �.dt
� �
p p

( )
a(
� �� p
I = a�
� � � �
� � �
t
�p � � �
� � �i =1 �
�1 � � �
i=
� � � �
p Ci t c +1.bi a p −i
�p �
� i c i p −i � p
Put n.a.i + a ( m + 1) − 1 =− I = a �� p t .b a �.dt = a.�
�C �

c
� � c +1
�i =1 �
� � i =1
� �

p Ci x n.i + m +1.bi a p −i p Ci x n.i + m +1.bi a p −i
( )
np p p
m
.dx = a.� =�
I = � . a + b.x
x
a ( n.i + m + 1) ( n.i + m + 1)
i =1 i =1
1
n.i + m +1 i p −i �
Ci .bi a p−i
� Ci x
1 p p
( )
.b a
np �p p
m
.dx = � �= �
� I1 = � . a + b.x
x
� 1 ( n.i + m + 1) � i =1 ( n.i + m + 1)
i=
� �
0 0




10
If n, m < 0, n.i + m + 1 < 0 � n.i + m + 1 = − f
c
� Ci x n.i + m +1.bi .a p−i �
c p
( )
np �p
.dx = lim =
m
=x �
I1 = lim . a + b.x
( n.i + m + 1)
c +1 � 1 �
c+ +−
. i=
� �
1 1
c c
� Ci x − f .bi .a p−i � Ci .bi .a p−i �

p p
�p p
� = lim
= lim � �f
� �
� ( n.i + m + 1) � c +1 � 1 x ( n.i + m + 1) �
c− +− i =1
. i=
� �
� � 1
1
p Ci .bi .a p −i p Ci .bi .a p−i
p p
= −�
( n.i + m + 1) i�n.i + m + 1
=
i =1 =1
( because n.i + m + 1 < 0 )

p
1
m n�
1 n �� �
( )
p
i i
��
+�� . � + b. � � � = + m . a + b.x n .dx
* lim a x
n n +i n i =1 � � � � �� 0
n n
� �
Ci .bi a p−i
p
p
( m, n, p is interger > 0 )
=+
i =1 ( n.i + m + 1)

( )
p 1 i
> 0 ) , xi � 0, 1] , ∆x =
[
* Put f ( x i ) = xi m . a + b.xi n ( m, n, p is interger , xi =
n n
p 1
m n
1 n �� � � ��
( )
n p
i i
. � + b. � � � = lim � ( x i ) .∆x = � . a + b.x n .dx
xm
�n�
� I = lim a f

n n +i n i =1 � � � � � � n +a i =1
n
m
� � 0
Ci .bi a p−i
p
p
=�
i =1 ( n.i + m + 1)




11
p Ci .bi a p −i
)
(
n
n1 p p
m m +1
. a + b. ( i.h )
� �
=−
* lim i h
i =1 ( n1.i + m + 1)
n + +i i =1
h = +i
p Ci .bi a p −i
p
( with n1.i + m + 1 < 0 )
=�
i =1 n1.i + m + 1

( )
p
( m, n1, p is intergers, p > 0, m, n1 0
x +1 − x −1 x +1 − x −1
2 2 4
� x + 1 − x − 1 = � 2 x + 1 = u + � 4 ( x + 1) = u 2 + 4 +
u2
u u
� 8�
� 4dx = � −
2u du
3�
� u�

8 4 4
� �
2u − u− � u4 �
4
1 u −4
3 3 2
1 1
I = � u du = � u du = � du − � u
du = �� du �
4 ( u + 2) 2 u+2 2 u 4 + 2u 3 2 � u 4 + 2u 3 u 2 + 2u �
� �
� �
4
2 2 2.dk 2
I1 = − � u du put u = � du = − ,k=
u 2 + 2u k2
k u
k2 k2
� 2.dk � 1
� I1 = − � − =� dk
4 � k2 � 2 ( 1 + k )
4
+� �
2k
k
− ( 2u + 2 )
u2 u 2 + 2u − 2u + 2 − 2 2
I2 = = =1+ +
u 2 + 2u u 2 + 2u u 2 + 2u u 2 + 2u

1 � u2 k2 �
1
�I= � du + �
� dk �
2 (1+ k) �
2 � u 2 + 2u
� �
d(1+ k) �
� 1 �k 2 − 1
2u + 2
1� 2
= �du − � du + � +� dk + �
� u 2 + 2u du � � �
4 � 1+ k 1+ k �
2
2� u + 2u � � �
� �
� �1
1 1
du � �( k − 1) dk + ln 1 + k �
2
= � − ln u + 2u + 2.� +�
u
2�2 � 4� �

2�
1+ �.u

� �
� u�
� �




15
2�

d � + 1�
−2du
2 1 2
u�

I3 = � du = − � = −� = − ln 1 +
� 2 � u2
� 2�2 � 2� u
1+ � �+ � �+ �
.u 1 1

� u� � u� � u�
2 � 1 �2 �
1� k
� I = � − ln u 2 + 2u − ln 1 + � � − k + ln 1 + k � C
+ +
u
2� u � 4� 2 �
2
Where 0 < u = = x + 1 + x − 1
k


( ) +2
2 d x2 + 1
2 2
2x + 1 dx 1 3
* I=� dx = � =− +J= +J
�2
( x 2 + 1) 1
( ) ( ) ( )
2 2 2 10
2 1 x2 + 1
1 x +1 1 x +1

π
π
2
( )
� =1� t =
x
dx
Put x = tan ( t ) � dx = 1 + tan ( t ) dt � �
2
J=� 4
( )
2
� = 2 � t = arctan ( 2 )
1 x2 + 1 +x


( ) = arctan( 2) cos2 ( t ) dt
arctan ( 2 ) 1 + tan 2 ( t ) dt
�J= � �
( tan 2 ( t ) + 1)
2
π /4 π /4

arctan ( 2 )
arctan ( 2 )
1 + cos ( 2t ) � sin ( 2t ) �
t
d ( 2t ) = � +
= � �
4 2 4�

π /4 π /4
sin ( 2.arctan ( 2 ) )
arctan ( 2 ) π1
= + −

2 4 84
1 arctan ( 2 ) sin ( 2.arctan ( 2 ) ) π
So I = + + −
20 2 4 8




16
4
dx
* I=+
x+ x
1
4
� =1� u =1
x
dx dx
* I=+ put x = u � = du � �
+x = 4 � u = 2
x+ x 2x
1
4 2 2 2
2u + 1
x 2u 1
� I = 2.� dx = � du = � du − � du
( ) u2 + u u2 + u u2 + u
x + x .2 x
1 1 1 1

( ) = ln u 2 + u 2 ,
2 d u2 + u
2
2u + 1
I1 = � du = �
u2 + u 2
1 u +u
1
1
du
� 1�
2− 2 2 d �+
1

2 2
1 u = � u � Ln 1 + 1
I2 = − � du = � =
1�1
u2 + u u1
11+ 1 1+
1
u u
12 3 3 9
2 2
+ I = Ln u + u 1 + Ln 1 + du = Ln6 − ln 2 + Ln − ln 2 = ln 3 + ln = Ln
u1 2 4 4


� 3�
3
2�2 + 1 − x+
x x�
dx
x2 + 1 2 2�
1 �
I=� =�
( )( )
x 4 − x 2 + 1 2 x 2 + 1 − 3.x x 2 + 1 + 3.x

( ) ( )
x 2 + 1 − 3x dx x 2 + 1 + 3x dx
1 1
=� +�
( )( ) ( )( )
2 x 2 + 1 − 3.x x 2 + 1 + 3.x 2 x 2 + 1 − 3.x x 2 + 1 + 3.x

1 dx 1 dx
= +�
�2
( ) ( )
2 x 2 + 1 − 3.x
2 x + 1 + 3.x




17
1 dx 1 dx
= +�
� 2 2
2� 2 2� 2
3� �� 3� ��
1 1
�+ �+ � � �− �+ � �
x x
2 � �� 2 2 � ��2
� �
� �
� 3� � 3�
�+ �−
x x
� �
� �
1 2 � arctg � 2 � + C
= .2 � �
+
arctg �
�1 � �1 �
2� �
�2 � �2 �
� �
� � � �
� �
( ) ( )
= arctg 2x + 3 + arctg 2x − 3 + C


x
arcsin x x. 1 − x 2
2 2
* I = � − u du = � − x dx = + +C
1 1
2 2
0

We have: I = The area of plane figure bounded by the graph y = 1 − x 2 and Ox.
And y = 1 − x 2 � y 2 + x 2 = 1 ( choose y > 0 ) is a half circle with radius R = 1.
We have the follow figure:




π
� �
We have: x = cos � − α � sin α � α = arcsin x
=
2
� �
I = area of yellow part = area of sector AOB + area of triangle BOC




18
α x. 1 − x 2 arcsin x x. 1 − x 2
I= + = +
2 2 2 2
x
arcsin x x. 1 − x 2
2 2
� I = � − u du = � − x dx = + +C
1 1
2 2
0




( x 2 − a 2 ) + a.ln
x−a x−a + x+a
* I=− dx = −
x+a x−a − x+a

( ) ( )
x−a x−a
� t 2 ( a + x ) = x − a � x t 2 − 1 = −a 1 + t 2
* I=− put t =
dx
x+a x+a

( ) � dx = ( )( )( )( ) dt
' '
−a 1 + t 2 −a 1 + t 2 t2 − 1 + a 1 + t2 t2 − 1
�x=
( t 2 − 1) ( t 2 − 1)
2


−2at ( t 2 − 1) + 2at ( 1 + t 2 ) 4a.t.dt 2t.dt
� I = a−
= dt = 2t.
( t 2 − 1) ( t 2 − 1) ( t 2 − 1)
2 2 2


d ( t 2 − 1)
2t.dt 2t.dt 1
put u = 2t, dv = �v=� =� =− , du = 2dt,
( t 2 − 1) ( t 2 − 1) ( t − 1)
( t 2 − 1)
2 2 2 2



t +1
2at dt 2at 1
+ 2a −
�I=− =− + 2a. .ln
( t 2 − 1) ( t 2 − 1) ( ) t −1
t2 − 1 2

x−a x−a x−a
+ 1 2a
2a
x−a x−a + x+a
x + a + a.ln x+a x + a + a.ln
�I=− dx = − =
� −a � −
x+a x � 2a �
x−a x−a − x+a
− 1� −1
� � �
� +a � � +a�
x x
x+a

( x 2 − a 2 ) + a.ln
x −a x+a x−a + x+a x −a + x +a
= −a + a.ln =−
.
x+a a x−a − x+a x−a − x+a

The second method:




19
x−a 1 x
I== put x = a.cos 2t � dx = −2a.sin2t.dt � t = .arccos
dx
x+a 2 a

))
((
x − a = a ( cos 2t − 1) = a 1 − 2sin 2 t − 1 = −2a.sin 2 t

x + a = a ( 1 + cos 2t ) = a ( 1 + ( 2 cos t − 1) ) = 2a.cos
2 2
t

−2a.sin 2 t i.sin t
� I = −2a � sin2t.dt = −2a � .2sin t.cos t.dt
2a.cos 2 t cos t
1 − cos 2t
= −4a.i � 2 t.dt = −4ai � dt = −2ai � + 2ai � 2t.dt = −2ai.t + ai.sin2t
sin dt cos
2
x ai 2x �

= −ai.arccos + .sin � arccos �
a2 a�


a−x � x�
1 2x �

�I=− dx = a.i � .sin � −
arccos � arccos �
a+x 2 a� a�


x −a + x +a
= − x 2 − a 2 + a.ln
x−a − x+a


�a−x �
a−x
( a − x) ( a + x)
* I=� dx = − 2a.arctan � �
x+a �a+x�

( )( )
a−x a−x
� t2 ( a + x ) = a − x � x t2 + 1 = a 1 − t2
* I=+ put t =
dx
x+a x+a

a ( 1 − t2 ) a ( 1 − t2 ) ( 1 + t2 ) − a ( 1 − t2 ) ( 1 + t2 )
' '

�x= � dx = dt
(1+ t ) (1+ t )
22
2


−2at ( 1 + t 2 ) − 2at ( 1 − t 2 ) −4a.t.dt 2t.dt
� I = −a +
= dt = 2t.
( 1 + t2 ) ( 1 + t2 ) ( 1 + t2 )
2 2 2


d ( 1 + t2 )
2t.dt 2t.dt 1
put u = 2t, dv = �v=� =� =− , du = 2dt
(1+ t ) (1+ t ) (1+ t )
(1+ t )22 22 22 2




20
�−2t � −2dt 2at
� ( −a ) � 2 =
� I = −a � − − 2a.arctgt
�+ t2 � 1 + t2
1+ t
1
a−x a−x
2a 2a
a + x − 2a.arctg � a − x � a + x − 2a.arctg � a − x �
a−x
�I=� dx = =
� � � �
a−x 2a
a+x �a+x � �a+x �
1+
a+x a+x
�a−x � �a−x �
a−x
( a + x ) − 2a.arctg � � ( a − x ) ( a + x ) − 2a.arctg �
= = �
a+x �a+x� �a+x �

a−x
The second method: I = + dx
a+x
1 x
put x = a.cos 2t � dx = −2a.sin2t.dt � t = .arccos
2 a

) ) = 2a.sin t
((
a − x = a ( 1 − cos 2t ) = a 1 − 1 − 2sin 2 t 2


a + x = a ( 1 + cos 2t ) = a ( 1 + ( 2 cos t − 1) ) = 2a.cos t
2 2



2a.sin 2 t sin t
� I = −2a � sin2t.dt = −2a � .2sin t.cos t.dt
2a.cos 2 t cos t
1 − cos 2t
= −4a � 2 t.dt = −4a � dt = −2a � + 2a � 2t.dt
sin dt cos
2
xa 2x �

= −2at + a.sin2t = −a.arccos + .sin � arccos �
a2 a�

�a−x �
a−x xa 2x �

arccos � a 2 − x 2 − 2a.arctg �
�I=� dx = −a.arccos + .sin � = �
a+x a+x�
a2 a�
� �




21
2001 � 2
1 �x � 1 �
x 2000 .dx�x � 1 �x �
* I=� =� − +
� �
� � � � �
�2003 � + 1 � 1001 � + 1 � 2001 �
( x + 1) 2004 � + 1 �
x x x
� �
2000 2
x 2000 .dx �−1� � 1 �� �
t 1
* I=� put t = x + 1 � dx = dt � I = � � − .dt
� ��2 �

( x + 1) 2004 �t � � t �� �
t
2000 2
�−1�
t � 1� � 1� 1 � 1� 1 x
=� � � �d � � dat u = 1 − � du = d � � u = 1 −
− − −� =

x +1 x +1
�t � �t� �t� t �t�
u 2003 2.u 2002 u 2001
( )
2
( u − 1) .du = �
2000 2002 2001 2000
�I=� − 2u +u .du = − +
u u
2003 2002 2001
2001 2
�x � � 1 �x � 1 �x � 1 �
x 2000 .dx
�I=� = �− +
� �
2004 � + 1 � � � � �
2003 � + 1 � 1001 � + 1 � 2001 �
( x + 1) x� x x
� � �


a +1 � 2
2 �x � 1 �
x a .dx �x � 1 �x �
* I=� =� − +
� �
� � + 3 � +1� a + 2 � +1� a +1�
( x + 1) a + 4 � + 1 �
x a x� x�
� �
� �
a 2
x a .dx � − 1 �� 1 �� �
t 1
* I=� put t = x + 1 � dx = dt � I = � �� �� �
− .dt

( x + 1) a + 4 � t �� t ��2 �
t
a 2
� − 1 �� 1 � � 1 �
t 1 � 1� 1 x
= � �� �d � � dat u = 1 − � du = d � � u = 1 −
− − −� =

x +1 x +1
� t �� t � � t � t �t�
u a +3 2.u a + 2 u a +1
( )
� I = � ( u − 1) .du = �a + 2 − 2u a +1 + u a .du =
2
ua − +
u
a +3 a+2 a +1
a +1 2
�x � �1 �x � 2 �x � 1 �
x a .dx
�I=� = − +
� �
a +4 � + 1 � � + 3 � + 1 � a + 2 � + 1 � a + 1 �
( x + 1) x�a x� x�
� � �
� �




22
dx
* I=+
ax 2 + bx + c
dx 1 dx
* I=� =�
ax 2 + bx + c a � b � c b2
2
� + �+ − 2
x
� 2a � a 4a

( )
1 dx
if b2 − 4ac > 0
=�
a� �� �
2 2
� + b + b − 4ac �x + b − b − 4ac �

x
� �� �
2a 2a
� �� �
4a 2 .dx
1
=�
a� �� �
2 2
� + b + b − 4ac �2a.x + b − b − 4ac �
2a.x �
� �� �

dx
*
� + b + b2 − 4ac �2a.x + b − b2 − 4ac �

2a.x
� �� �
� �� �
A B
= +
� + b + b2 − 4ac � � + b − b 2 − 4ac �
2a.x 2a.x
� �� �
� �� �
� 1 = A � + b − b2 − 4ac � B � + b + b 2 − 4ac �
+ 2a.x
2a.x
� �� �
� �� �
� A + B = 0 � A = −B, b ( A + B ) + b 2 − 4ac ( B − A ) = 1 � 2B b 2 − 4ac = 1
1 1
�B= , A=−
2 b2 − 4ac 2 b 2 − 4ac

4a.dx
�I=�
� + b + b 2 − 4ac �2a.x + b − b 2 − 4ac �

2a.x
� � � �
� � � �
4a.dx 4a.dx
= −� +�
2 b2 − 4ac � + b + b 2 − 4ac � 2 b2 − 4ac � + b − b 2 − 4ac �
2a.x 2a.x
� � � �
� � � �
2a.d � + b + b 2 − 4ac � 2a.d � + b − b2 − 4ac �
2a.x 2a.x
� � � �
� � � �
2a 2a
= −� +�
b2 − 4ac � + b + b 2 − 4ac � b 2 − 4ac �a.x + b − b 2 − 4ac �
2a.x 2
� � � �
� � � �


23
− ln � + b + b 2 − 4ac � ln � + b − b 2 − 4ac �
+
2a.x 2a.x
� � � �
� � � �
=
b 2 − 4ac

2a.x + b − b 2 − 4ac
dx 1
�I=− = ln
ax 2 + bx + c b2 − 4ac 2a.x + b + b 2 − 4ac

� �
� �
2a 1 1
− +
Check the solution : � �
� � + b + b 2 − 4ac � � + b − b 2 − 4ac �
2 �
b − 4ac � �2a.x 2a.x
�� ��
�� �� ��
�� �
�� �
2 2
� � + b − b − 4ac � � + b + b − 4ac �
− 2a.x + 2a.x �
2a � �� �
= � �
)
( 2
( 2a.x + b ) − b + 4ac
2
2
b − 4ac � �
� �
� �
� �
2 b 2 − 4ac
� �
2a 4a 1
= = =
)(
( )
� �
ax 2 + bx + c
2 22
b 2 − 4ac �( 2a.x + b ) − b + 4ac � 4a .x + 4ab.x + 4ac
2
� �

dx 1 dx
* I=� =�
ax 2 + bx + c a � b � c b2 2
� + �+ − 2
x
� 2a � a 4a

( )
1 dx
if b 2 − 4ac < 0
=�
2
a
b � � 4ac − b2 �
2

� + �+ � �
x
2a � � 2a �

� �
� �
� x+ b � � 2ax + b �
1 2a 2
� 2a �
=. = .arctan � �
.arctan
� 2� � �
a 4ac − b 2 2 2
� 4ac − b � 4ac − b � 4ac − b �
� �
� 2a �




24
� b�
d� + �
x
( if b2 − 4ac = 0)
dx 1 dx 1 � 2a �
* I=� =� =�
2 2 2
a � b � c b2 a � b�
ax + bx + c
� + �+ − 2 �+ �
x x
� 2a � a 4a � 2a �
−2
1
�I=− =
� b � 2ax + b
a� + �
x
� 2a �


dx
* I=+
( )
2
ax 2 + bx + c
2
1
� �

�� �
b 2 − 4ac �
� 2a.x + b + �
( 2a ) � � �
2
dx � ��dx
* I=� = �
( )
b2 − 4ac � �
2 1
ax 2 + bx + c +
� �
�� +b− �
2 �
b − 4ac �
2a.x
�� �
�� ��

( if )
b 2 − 4ac > 0


�� �
d � + b + b2 − 4ac � d � + b − b 2 − 4ac � �
2a.x 2a.x
� � �
�� �+ � �
� �
� �
2 2
2�� �
2a � + b + b 2 − 4ac � 2a � + b − b 2 − 4ac � �
4a 2a.x 2a.x
� � �
��
= � � ��
b2 − 4ac �
2dx
� �
−�
)
(
� �
2
� ( 2a.x + b ) − b + 4ac
2

� �




25
� �
� �
2
−4a 1 1
= +
� �
( )
2 � + b + b 2 − 4ac � � + b − b 2 − 4ac �
2a b − 4ac � 2ax �
2ax
� �� �
� �
� �� �
� �
4a 2 2dx
− �
( 4a 2 .x 2 + 4ab.x + 4ac)
b2 − 4ac

� �
4a 2
−2a 4ax + 2b dx
� �
= − �
( )( ) ( ) ( a.x 2 + b.x + c)
� �
b 2 − 4ac �4a 2 .x 2 + 4ab.x + 4ac � 2a b2 − 4ac
� �
� �
−2a 2ax + b 2a dx
� �

= −
( )( ) ( b2 − 4ac) ( a.x 2 + b.x + c )
�2 �
2
2a b − 4ac �a.x + b.x + c �
� �



2a.x + b − b 2 − 4ac
dx 2
* I1 = − = .ln
ax 2 + bx + c b2 − 4ac 2a.x + b + b 2 − 4ac
dx
�I=−
( )
2
ax 2 + bx + c

� �
2a.x + b − b 2 − 4ac
−1
� 2ax + b 2a

= − .ln
( )( )
� �
( )
b 2 − 4ac �a.x 2 + b.x + c 3
2a.x + b + b2 − 4ac

b 2 − 4ac
� �

� �
2a.x + b − b 2 − 4ac
−1
� 2ax + b 2a �
= + .ln
( )( )
�2 �
2 2
2a.x + b + b 2 − 4ac
b − 4ac �a.x + b.x + c �
b − 4ac
� �

�1 �
2
dx
�1 1� 1 2
� �
* I=� =� − dx = � + − dx
� �
�x + 1) 2 ( x + 2 ) 2 ( x + 1) ( x + 2 )
( )
�+ x 2 + x �
2
( �
1
2
x + 3x + 2 � �

� 2x + 3 �
1 1 �1 1�
=− − − 2� − � = −� 2 � 2ln x + 1 + 2ln x + 2

dx

x +1 x + 2 � +1 x + 2 �
x � + 3x + 2 �
x
� 2x + 3 � x+2
= −� +
� 2 ln
x +1
� 2 + 3x + 2 �
x

26
mx + n
* I=+ dx
2
ax + bx + c

( where 4ac − b2 > 0)
mx + n
2 / Evaluate + dx
ax 2 + bx + c
2
� 2 bx c � �� b � 4ac − b �
2
2
ax + bx + c = a � + + � a �x + � +
=� �
x
a a� � 4a 2 �
� 2a �
� � �
4ac − b2
( if 4ac − b2 > 0)
b
e2 =
put : u = x + ,
2
2a 4a
� b�
m� − � n +
u
mx + n 1 � 2a � du = m 2udu + 1 � − m.b � du
+
dx = �
��
2a � + e2 a � 2a � u 2 + e2
n
� �
ax 2 + bx + c u 2 + e2 u2
a

( ) ( )
d u 2 + e2 m.ln u 2 + e2
m 1 � m.b � 1 u 1 � m.b � u
+ u 2 + e2 + a � − 2a � arctg e =
= + �−
n n arctg

2a e 2a a.e � 2a � e
� �


( )
m.ln u 2 + e2
mx + n 1 � m.b � u
�I=� dx = + n− arctg
� �
ax 2 + bx + c 2a a.e � 2a � e
� 2 + bx + c � � �
ax
� x+ b
m.ln � � �
� �
a � − mb �
1 2an
� � arctg � �
2a
= + � � � �
2a a 4ac − b2 � 2a � 2
� 4ac − b �
� �
� 2a �
2a


( )
m.ln ax 2 + bx + c − m.ln a � 2an − mb � � 2ax + b �
= +� +

.arctg � �C
� � � �
2a 2 2
� 4ac − b � 4ac − b
a � �

( ) + �2an − mb
m.ln ax 2 + bx + c � � 2ax + b �
= +
��
.arctg � �C
� 2� � �
2a 2
� 4ac − b � � 4ac − b
a �
m.ln a
� �
Because − is constant �

2a
� �




27
�3 �
1� �
x . x 2 + a 2 + � x 2 + a 2 + a.ln x + x 2 + a 2 �
x.
1� �
2� �+ C
* I = � x 2 + a 2 .dx = �
x2 �
4� 3 �
x

�a .arctan a �
� �

()
1 1� �
* I = � x 2 + a 2 dx = � 2 + a 2 .d x 3 = �3 . x 2 + a 2 − � .d � x 2 + a 2 �
x2 x3 �
x x ��
� �
3 3� �
1 �3 � 1� �
x4 x4 − a4 + a4
2 2 3 2 2
= � . x +a −� .dx � � . x + a − �
=x .dx �
x
3� � 3�
2 2 2 2 �
x +a x +a
� �� �

( )( )
� �
x2 − a2 x2 + a2 a4
1 �3
.dx �
2 2
= x . x +a −� .dx − �
� �
3 1 + x2 x2 + a2
� �
� �

( )
1� 1 x�
= �3 . x 2 + a 2 − � 2 − a 2 x 2 + a 2 .dx − a 4 . .arctan �
x x
3� a a�
1� x�
= �3 . x 2 + a 2 − � x 2 + a 2 .dx + � 2 + a 2 .dx − a 3 .arctan �
x2
x x
3� a�
1� x�
I = �3 . x 2 + a 2 − I + � 2 + a 2 dx − a 3 .arctan �
x x
3� a�
4 1� x�
� .I = �3 . x 2 + a 2 + � 2 + a 2 .dx − a 3 .arctan �
x x
3 3� a�
1 �3 x�
x . x 2 + a 2 + � 2 + a 2 .dx − a 3 .arctan � C
I= +
x
4� a�


xu = x 2 + a 2 x.dx
x
2 2
* I1 = = x + a dx put � � du = , v=x
2 2
dv = dx
= x +a
+
x 2 .dx x2 + a − a
� I = x. x 2 + a 2 − � = x. x 2 + a 2 − � dx
2 2 2 2
x +a x +a
a.dx
I = x. x 2 + a 2 − � 2 + a 2 dx + �
x
x2 + a2
= x. x 2 + a 2 − I + a.ln x + x 2 + a 2




28
� 2I = x. x 2 + a 2 + a.ln x + x 2 + a 2

1� �
� I = � x 2 + a 2 + a.ln x + x 2 + a 2
x. �
2� �
1� x�
� I = � x 2 + a 2 .dx = �3 . x 2 + a 2
x2 + � 2 + a 2 .dx − a 3 .arctan �
x x
4� a�

� I = � x 2 + a 2 .dx
x2
1 �3 x�
2 1� 2� 3
2
x. 2 2 2
= � . x + a + 2 � x + a + a.ln x + x + a � a .arctan a �

x
4� � � �

a
� J = � x 2 + a 2 .dx
x2
0

( )
1� 1 a �
= �3 . 2.a + a 2 . 2 + a.ln a + a 2 − a 3 .arctan1 − .ln a �
a
4� 2 2 �
2a 3 .π
1� �
( )
4 2
= � 2.a + a . 2 + a.ln a 1 + 2 − − a.ln a �
2.
8� �
4
� �
a 3 .π �
1� 4 2
= � 2.a + a . 2 + a.ln 1 + 2 − �
2.
8� 2�
� �


� ax ( n + 1) − b �
n +1
n
* I = � ax + b ) dx = ( ax +2 )
x( � (n −1, − 2 )

b
� 2 ( n + 1) ( n + 2 ) �
a
� �
n n
* I = + ( ax + b ) dx dat u = x, dv = ( ax + b ) dx
x

( ax + b ) n +1
n
� du = dx, v = + ax + b ) dx =
(
a ( n + 1)
n +1
( ax + b ) n +1 .dx
x ( ax + b )
−+
�I=
a ( n + 1) a ( n + 1)




29
n +1
( ax + b ) n + 2
x ( ax + b )
= −
a ( n + 1) a 2 ( n + 1) ( n + 2 )

n +1 ax ( n + 2 ) − ( ax + b )
� �
= ( ax + b ) � �
� a ( n + 1) ( n + 2 ) �
�2 �

� ax ( n + 1) − b �
n +1
= ( ax ++ ) � (n −1, − 2 )

b
� 2 ( n + 1) ( n + 2 ) �
a
� �

−1 −1
* I = + ( ax + b ) dat u = x, dv = ( ax + b )
x dx dx
ln ax + b
−1
� du = dx, v = + ax + b )
( dx =
a
x.ln ax + b ln ax + b .dx
−+
�I=
a a
x.ln ax + b 1
+ln ax + b .d ( ax + b )
= −
2
a a
a.dx
put u = ln ax + b , dv = d ( ax + b ) � du = , v = ax + b
ax + b

1
+ln ax + b .d ( ax + b )
� I1 = −
2
a
( ax + b ) .ln ax + b a ( ax + b ) dx ( ax + b ) .ln ax + b
1 x
+
=− + =− +
ax + b
a2 a2 a2 a
( ax + b ) .ln ax + b
x.ln ax + b x
�I= − +
a2
a a
b.ln ax + b
x.dx x
�I=+ =−
( ax + b ) a a2

1
ln ax + b .d ( ax + b )
* I = � ax + b .dx =
a�
ln

a.dx
put u = ln ax + b , dv = d ( ax + b ) � du = , v = ax + b
ax + b
ln ax + b ( ax + b ) 1 a ( ax + b ) .dx ln ax + b ( ax + b ) − ax
�I= −� =
ax + b
a a a



30
−2 −2
I = + ( ax + b ) put u = x, dv = ( ax + b ) dx � du = dx,
x dx
1 1
( ax + b ) −2 d ( ax + b ) = −
a+
v=
a ( ax + b )
d ( ax + b ) ln ax + b
x dx x x
�I=− +� =− +� =− +
a ( ax + b ) a ( ax + b ) a ( ax + b ) a ( ax + b )
a 2 ( ax + b ) a2


−ax ( n − 1) − b
x.dx
(n 2)
* I=+ =−
a 1, n
( ax + b ) n a ( n − 1) ( n − 2 ) ( ax + b ) n −1
x.dx dx
* I=− put u = x, dv =
( ax + b ) n ( ax + b ) n
( ax + b ) 1−n ( ax + b ) 1−n
dx
� du = dx, v = − = =−
1− n n −1
n
( ax + b )
1− n
( ax + b ) 1−n .d ( ax + b )
− x ( ax + b )
+−
�I=
a ( n − 1)
n −1

1− n
( ax + b ) 2−n 1− n 2− n
− x ( ax + b ) −ax ( 2 − n ) ( ax + b ) + ( ax + b )
= + =
a ( n − 1) ( 2 − n ) a ( n − 1) ( 2 − n )
n −1

( ax + b ) 1−n �ax ( 2 − n ) + ( ax + b ) �

� �
=
a ( n − 1) ( 2 − n )
ax ( n − 1) + b −ax ( n − 1) − b
(n 2)
I= =−
− 1, n
n −1 n −1
a ( n − 1) ( 2 − n ) ( ax + b ) a ( n − 1) ( n − 2 ) ( ax + b )




31
+x
( n − 3) !.m!
x m +1.dx
(n 3)
+ =+
*
n+m
a m +1. ( n + m − 1) !.bn −2
0 ( ax + b )
−ax ( n − 1) − b
x.dx
(n 3)
* I=+ =−
( ax + b ) n a ( n − 1) ( n − 2 ) ( ax + b ) n −1
+x
+x � �
ax ( n − 1) + b
x.dx
� J( a) = + = −� �
� ( n − 1) ( n − 2 ) ( ax + b ) n −1 �
n
0 ( ax + b ) a
� � 0
b 1
= =
a ( n − 1) ( n − 2 ) b n −1 a ( n − 1) ( n − 2 ) bn −2
+−
( −n ) x 2 .dx ( −1)
� J ( a) =
'
+ = ,
n +1 n −2
a . ( n − 1) ( n − 2 ) b
2
0 ( ax + b )

+.
( −n ) ( −n − 1) x3 .dx ( −1) ( −2 )
J ( a) =
''
+ =
( ax + b ) n + 2 a 3 . ( n − 1) ( n − 2 ) b n −2
0
+−
( −1) m ( n + m − 1) !.x m+1.dx ( −1) m .m!
J( a)( ) =
m
+ =
n +m
a m +1. ( n − 1) ( n − 2 ) bn −2
( n − 1) !.( ax + b )
0
++
( n − 1) !.m!
x m +1.dx
�+ =
n +m
a m +1. ( n − 1) ( n − 2 ) ( n + m − 1) !.bn −2
0 ( ax + b )
( n − 3) !.m!
=
a m +1. ( n + m − 1) !.b n −2

+x
+x � �
ax ( n − 1) + b
x.dx
J ( b) = + � �
=−
n −1 �
n
0 ( ax + b )
� ( n − 1) ( n − 2 ) ( ax + b )
a
� � 0
b 1
= =
a ( n − 1) ( n − 2 ) b n −1 a ( n − 1) ( n − 2 ) bn −2
+−
( −n ) .x.dx ( −1) ( n − 2 )
� J ( b) =
'
+ = ,
n +1
a. ( n − 1) ( n − 2 ) bn −1
0 ( ax + b )
+−
( −1) 2 .n ( n + 1) .x.dx ( −1) 2 ( n − 2 ) ( n − 1)
J ( b) =
''
+ =
n +2
a 3 . ( n − 1) ( n − 2 ) b n
( ax + b )
0


32
+.
( −1) m ( n + m − 1) !.x.dx
J ( b) ( ) =
m
+ n+m
0 ( n − 1) !. ( ax + b )
m
m� �
( −1) . � ( n + i − 3) �
+
( −1) m .( n − 2 ) ( n − 1) ... ( n + m − 3) � �
�3 �
i=
= =
a. ( n − 1) ( n − 2 ) b n + m −2 a.bn + m −2
m
� �
( n − 1) !.� ( n + i − 3) �
+
+− � �
x.dx �3 �
i=
�+ =
n +m n + m −2
a. ( n + m − 1) !.b
0 ( ax + b )
( n + m − 3) ! 1
= =
a. ( n + m − 1) !.b n + m −2 a. ( n + m − 1) ( n + m − 2 ) .bn + m −2


x 2 .dx
* I=+
( ax + b )
x 2 .dx dx
put u = x 2 , dv =
* I=� � du = 2x.dx,
( ax + b ) ( ax + b )
1 d ( ax + b ) ln ax + b
dx
v=� =� =
( ax + b ) a ( ax + b ) a
x 2 ln ax + b x.ln ax + b .dx
�I= − 2�
a a

−2
a+
I1 = x.ln ax + b .dx put u = ln ax + b ,

x2
a.dx
dv = x.dx � du = ,v=
ax + b 2
−2.x 2 ln ax + b 2 x 2 .a.dx
++
� I1 =
a ax + b
2a
x 2 ln ax + b x 2 ln ax + b
2 x 2 .a.dx
++
�I= − � I = 0 + 2I
a ax + b
a a
So choosing u and v such that is wrong.




33
x 2 .dx x.dx
* I== put u = x, dv =
( ax + b ) ( ax + b )
x b.ln ax + b
x.dx
� du = dx, v = = =−
( ax + b ) a a2
ax 2 − x.b.ln ax + b � b.ln ax + b �
x
−= −
�I= dx
� �
a2 a2
a
� �
ax 2 − x.b.ln ax + b x2 b
2a a 2 =
= − + ln ax + b .dx
a2
b b
� ax + b .dx = a3 � ax + b .d ( ax + b )
I1 = ln ln
2
a

a.dx
put u = ln ax + b , dv = d ( ax + b ) � du = , v = ax + b
ax + b
b ( ax + b ) .ln ax + b b a ( ax + b ) dx b ( ax + b ) .ln ax + b − abx
� I1 = −� =
ax + b
a3 a3 a3
x 2 b ( ax + b ) .ln ax + b − abx
ax 2 − x.b.ln ax + b
�I= − +
a2 a3
2a
a 2 .x 2
1 �2 2 �
+ ax.b.ln ax + b + b2 .ln ax + b − abx �
= � .x − ax.b.ln ax + b −
a
a3 � �
2
� �
x 2 .dx 1 � 2 .x 2 �
a
− abx + b 2 .ln ax + b �
�I=� =�
( ax + b ) a 3 � 2 �
� �

x 2 .dx x.dx
* I=+ put u = x, dv = � du = dx,
2 2
( ax + b ) ( ax + b )
ln ax + b
x.dx x
v=+ =− +
a ( ax + b )
( ax + b ) 2 a2

x2 x.ln ax + b ln ax + b
� �
x
−+−
�I=− + + dx
� �
a ( ax + b ) � a ( ax + b ) �
a2 a2
� �
x2 x.ln ax + b x.dx 1
=− + +� − ln ax + b .dx
a ( ax + b ) a 2 �
a ( ax + b ) a2
x b.ln ax + b
1 x.dx
I1 = � = −
a ( ax + b ) a 2 a3
34
1 � ax + b ( ax + b ) − ax �
ln
1
I2 = − � ax + b .dx = −
ln � �
a2 a2 � a �
x b.ln ax + b ln ax + b ( ax + b ) − ax
x2 x.ln ax + b
�I=− + + − −
a ( ax + b ) a2 a2 a3 a3
x2 2b.ln ax + b 2x 1 � a 2 .x 2 �

=− − + =� − 2b.ln ax + b + 2ax �
a ( ax + b ) a 2 a 3 �ax + b )
( �
a3 � �

'
1 � a 2 .x 2 �

− 2b.ln ax + b + 2ax �
Check the solution : �
a 3 �ax + b )
( �
� �

( )
�2 �
1 � a . 2x ( ax + b ) − a.x
2
− 2ab
+ 2a �
=� − �
ax + b
a3 � ( ax + b ) 2 �
� �

( )
�2 �
2
1 � a . a.x + 2bx
− 2a ( ax + b ) �
2ab
=� − +
ax + b �
ax + b
3 2
a � ( ax + b ) �
� �
1 � a .x − 2a bx + 2a x ( ax + b ) �
−3 2 2 2
x2
� �
= =
a3 � 2 � ( ax + b ) 2
( ax + b )
� �

x 2 .dx x.dx
I=− put u = x, dv = � du = dx,
n n
( ax + b ) ( ax + b )
ax ( 1 − n ) − b
x.dx
v=− =
( ax + b ) n a ( n − 1) ( n − 2 ) ( ax + b ) n −1
� �
ax 2 ( 1 − n ) − bx ax ( 1 − n ) − b
n −1 −
−� �
�I= dx
� ( n − 1) ( n − 2 ) ( ax + b ) n −1 �
a ( n − 1) ( n − 2 ) ( ax + b ) a
� �
ax 2 ( 1 − n ) − bx ax ( n − 1) dx
+−
=
n −1 n −1
a ( n − 1) ( n − 2 ) ( ax + b ) a ( n − 1) ( n − 2 ) ( ax + b )
b.dx
+−
n −1
a ( n − 1) ( n − 2 ) ( ax + b )



35
ax 2 ( 1 − n ) − bx 1 x.dx

= +
( n − 2)
n −1
( ax + b ) n −1
a ( n − 1) ( n − 2 ) ( ax + b )
b dx

+
a ( n − 1) ( n − 2 ) ( ax + b ) n −1
ax ( 1 − ( n − 1) ) − b
1� �
1 x.dx
− � �
I1 = =
( n − 2) n −1 ( n − 2 ) � n −2 �
( ax + b ) � ( n − 2 ) ( n − 3) ( ax + b )
a �
−ax ( n − 2 ) − b
=
( n − 3) ( ax + b ) n −2
2
a ( n − 2)

d ( ax + b )
b dx b
I2 = =
� �
a ( n − 1) ( n − 2 ) ( ax + b ) n −1 ( ax + b ) n −1
a 2 ( n − 1) ( n − 2 )
−b
b
= =
n −2
( ax + b ) n −2
2
a 2 ( n − 1) ( n − 2 ) ( ax + b ) ( 2 − n) a 2 ( n − 1) ( n − 2 )
ax 2 ( 1 − n ) − bx −ax ( n − 2 ) − b
�I= +
n −1
( n − 3) ( ax + b ) n −2
2
a ( n − 1) ( n − 2 ) ( ax + b ) a ( n − 2)
−b
+
( ax + b ) n −2
2
a 2 ( n − 1) ( n − 2 )


x c .dx
* I=+
( )
b
xa + m

x c .dx
with condition : c = a − 1 + na or ( c + 1) Ma put t = x a + m
* I=�
( )
b
xa + m
n
x na .dt ( t − m ) .dt
x a −1+ na .dt
( xa = t − m )
a −1 c
� dt = a.x .dx � x .dx = = =
a −1 a a
a.x
( t − m ) n .dt
�I=�
a.t b




36
1
x 3dx
* I=+
2
0 x + x +1

x3 � − x 2 + 1 �
x
� �
1 1 1
3
x dx
dx = � � x 2 + 1 − x �
� � x3 �
* I=� =� dx

� + x 2 + 1 �x − x 2 + 1 �
� � �
2
0 x + x +1 0 �x �
� �0
� �
� �
1
1 1�5 �
x 1
3 2 4
= � x + 1dx − � dx = I1 − � � = I −
x x
5 5
��
��
0 0 0

x = 0 � t =1
put t = x 2 + 1 � t 2 = x 2 + 1 � 2tdt = 2xdx � tdt = xdx.
x =1� t = 2
2
1 2 2
�5 t 3 �
( ) ( )
t
2 2 2 4 2
I1 = � x + 1.x.dx = � − 1 t.tdt = �t − t dt = � − �
x t
5 3�

� �
0 1 1 1
� 2 2 2 � � 1� 2 2 2 2 2 −1
4 1
=� − ��− �
− = + . So I =
�5 3 � � 3 � 15 15
5 15



* I = = 2 a 2 − x 2 .dx
x
x
* I = � a 2 − x 2 .dx
x2 put x = a.sin t � dx = a.cos t.dt, t = arcsin
a
a3
( ) 2
� 2t )
( sin
2 2 2 2 3 2 2
� I = � .sin t a 1 − sin t .a cos t.dt = a � t.cos t.dt =
a sin .dt
4
a 3 1 − cos 4t a 3 � sin 4t �
= dt = �−
4� 2
t �
8� 4�
a3 � �
x1 � x�
2 2 2
�I=� a − x .dx = � arcsin − sin �4.arcsin � + C
x �
8� a4� a�





37
a 2 − x 2 .dx
* I==
x2
a 2 − x 2 .dx x
* I=� put x = a.sin t � dx = a.cos t.dt, t = arcsin
x2 a

( )
a 2 1 − sin 2 t .a cos t.dt
cos 2 t.dt
( )
= � g 2 t + 1 − 1 dt = − cot gt − t
�I=� =� cot
a 2 .sin 2 t sin 2 t
a 2 − x 2 .dx x� x

�I=� = − cot g � −
arcsin � arcsin
x2 a� a



a 2 − x 2 .dx
* I==
x
a 2 − x 2 .dx π π
* I== put : x = a.sin t, − � � � dx = a cos t.dt
t
x 2 2
a 2 − x 2 = a cos t =2 cos t, because cos t
a 0
cos 2 t 1 − sin 2 t � dt t

� I = a� dt = a � dt = a �� t − � t.dt � a.ln tan 2 + a cos t + C
=
sin
sin t sin t � sin �
t t
2sin 2
sin
2 = 1 − cos t , sin t = x ,
t 2=
tan =
2 cos t 2 cos t .sin t sin t a
2 2 2

x2 a2 − x2
x� x�
� 2�
cos t = cos �
arcsin � 1 − sin �
= arcsin � 1 −
= =
2
a� a� a
� � a

a 2 − x 2 .dx a − a2 − x2
+ a2 − x2 + C
�I=� = ln
x x




38
* I = = 3 a 2 − x 2 .dx
x

* I = � a 2 − x 2 .dx
x3 put t = a 2 − x 2 � t 2 = a 2 − x 2

� 2t.dt = −2x.dx, x 2 = a 2 − t 2
−a 2 .t 3 t 5
( )
2 2 2 2 2 4
� I = −� − t t .dt = −a �.dt + �.dt = + +C
a t t
3 5
3 5
−a . � a − x � � a 2 − x 2 �
2� 2 2�
� �
� �+ � �+ C
32 2
� I = � a − x .dx =
x
3 5


* I = = a 2 − x 2 .dx
x

* I = � a 2 − x 2 .dx put t = a 2 − x 2 � t 2 = a 2 − x 2
x

� 2t.dt = −2x.dx, x 2 = a 2 − t 2
3
�2 2�
3 −� a − x �
t
� I = − �.dt = − = � �
t2
3 3


* I = = 5 a 2 − x 2 .dx
x

* I = = 5 a 2 − x 2 .dx put t = a 2 − x 2
x

� t 2 = a 2 − x 2 � 2t.dt = −2x.dx, x 2 = a 2 − t 2

( ) ( )
22
� I = − �2 − t 2 t .dt = − �4 − 2a 2 .t 2 + t 4 t 2 .dt
a a

−a 4 .t 3 2a 2 .t 5 t 7
= + −
3 5 7
� I = � a 2 − x 2 .dx
x3
3 5 7
−a . � a − x � 2a � a − x � � a 2 − x 2 �
4� 2 2� 2� 2 2�
� �
� �+ � �− � �+C
=
3 5 7




39
dx
* I=+
x. x 2 + a 2
dx t.dt
put t 2 = x 2 + a 2 � 2t.dt = 2x.dx � dx = , x2 = t2 − a2
* I=+
x
x. x 2 + a 2
t−a
t.dt t.dt 1
�I=� =� = .ln +C
( ) t+a
2 2 2a
2 2 2 t −a t
x x +a

x2 + a2 − a
dx 1
�I=� = .ln +C
x. x 2 + a 2 2a 2 2
x +a +a

dx
−a.dx
2
dx a dx dt
=�x
I=� = t � dt = =−

put
x2 x2
x a
x. x 2 + a 2 a2
1+
x2
−1
1 dt 1 1
2
�I=− � 2 = a .ln t + t + 1 = a ln
a t + t2 + 1
t +1

a2 a 1 x2 + a2 − a
1 1
2
= ln t − t + 1 = ln 1 + − = ln +C
x2 x a
a a x


2
� x2 + a2 − a �
� �
x2 + a2 − a x2 + a2 − a
1 1 1 � �
= .ln = .ln
.ln
� x2 + a2 + a � x2 + a2 − a �

2a x2 + a2 + a a x2 + a2 + a a � �
� �
� �
� �

x2 + a2 − a x2 + a2 − a
1 1
= .ln = .ln
a a x
x2 + a2 − a2




40
x 2 .dx
* I=+
( )
2
x2 + a2

x 2 .dx
( ) x
put x = a.tan t � dx = a tan 2 t + 1 .dt, t = arctan
* I==
( )
2 a
x2 + a2

( ) ( )
a 3 .tan 2 t. tan 2 t + 1 .dt 2
1 tan 2 t.dt 1 tan t + 1 − 1 .dt
�I=� =� =
a tan 2 t + 1 a � tan 2 t + 1
( )
2
a 4 tan 2 t + 1


( )
1 � 1 + cos 2t �
1� �1
dt
= t − � 2 t.dt = � − �
= � dt − �
� tan 2 t + 1 � a cos t .dt �
a� a� 2 �

x 2 .dx
1 � t 1 sin 2t � t sin2t 1 x1 x�

�I=+
= �− − . = − = .arctan − .sin �
t 2arctan �

( )
2 2a
a � 2 2 2 � 2a 4a a 4a a�

x2 + a2


x
x� ax
� a
= =
sin �2arctan �
2 2 2
a�
� � � a +x
x
1+ � �
a
��
x 2 .dx 1 x1 x� 1 x1 ax

�I=+ = .arctan − .sin � = .arctan − .
2arctan �
( )
2 a 4a a 2 + x 2
2a a 4a a � 2a

x2 + a2
1 x x
= .arctan −
( )
a 4 a2 + x2
2a




41
z
* x = sin ( arctan z ) =
1 + z2
* y = arcsin x � x = sin y, z = tan y � y = arctan z � x = sin ( arctan z )
sin ( arcsin x ) x x
z = tan ( arcsin x ) = = =
cos ( arcsin x ) 1 − sin 2 ( arcsin x ) 1 − x2

z2
( ) =x ( ) =z
x 2 2 2 2 2 2 2
z= � z 1− x � x 1+ z �x =
1 + z2
2
1− x
z
� x = sin ( arctan z ) = just take the positive value
2
1+ z

1
* x = cos ( arctan y ) =
1 + y2

1 − cos 2 ( arccos x )
sin ( arccos x ) 1 − x2
* y = tan ( arccos x ) = = =
cos ( arccos x ) x x

1 − x2
( ) ( )
1
� y2 .x 2 = 1 − x 2 � x 2 y 2 + 1 = 1 � x 2 =
y=
1 + y2
x
1
� x = cos ( arctan y ) = chi lay giá tri duong
2
1+ y
x 1 x
� sin ( 2 arctan x ) = 2sin ( arctan x ) .cos ( arctan x ) = =
.
1 + x2
1 + x2 1 + x2



'
'
x� 1 � � �
�1 x�
* Check the solution : � .arctan �− . � � arctan �
sin 2 �
2a a � 4a � � a�
� �
1 1
1 2 x�

a a
=. −. .cos � arctan �
2
2 2
2a � � 4a � � a�

x x
+1 +1
�� ��
a a
�� ��
1 1 1 1 x�

=. −. .cos � arctan �
2
2 x2 + a2 2 x2 + a2 a�

+� x� +
1 1 1
= −. +
. � � arctan � 1�
cos 2
x2 + a2 2 x2 + a2 � � a� �
42
1 1 x�

.cos 2 �
= − arctan �
x2 + a2 x2 + a2 a�

� �
� �
� �
� �
1 1 1 1
. �− �
= . �− =

1 1
2� 2�
2 2� 2
x�
2� �
x +a arctan � x + a � 1 + � � �
x
� 1 + tan � � � ��

a�

� � a
� ��

a2 � x2

1
= . �− =

1
( )
x2 + a2 � x2 + a2 � x2 + a2 2
� �



x 3 .dx
* I=+
( )
n
x2 + a2

x 3 .dx
put t = x 2 + a 2 � dt = 2x.dx, x 2 = t − a 2
* I=+
( x2 + a ) 2n


( t − a 2 ) .dt = a 2 dt 1 1−n a2
dt
�I= �n −1 − 2 � = 2 � .dt + 2 n − 1 t n −1
� t
( )
2t n n
2t t
a2
1
= +
2 ( 2 − n ) t n −2 2 ( n − 1) t n −1
x 3 .dx a2
1
�I=� = +
( ) ( ) ( )
n −2 n −1
2
2 ( 2 − n ) x2 + a2 2 ( n − 1) x 2 + a 2
x2 + a2




43
x 2m +1.dx
* I =a+ m, n 0
( x2 + a )
2n


x 2m +1.dx
put t = x 2 + a 2 � dt = 2x.dx, x 2 = t − a 2
* I=+ m, n �0
( )
n
x2 + a2


()
m m−k
(t −a ) + Ck .t k . a 2
2m .dt
m
.dt
()
1 m k k −n 2 m−k
k =0
+
�I=� =� = � Cm .t .a .dt
2t n 2t n 2 k =0

() ( ) ()
k k − n +1 2 m − k 2 k − n +1 m−k
k 2
. a2
m Cm . x + a
m C m .t .a m +1
1 x .dx 1
� = .�
= �I=�
( k − n + 1) ( k − n + 1)
( )
2n
2 k =0 2 k =0
2
x +a

�−1 i n −1−i �
n
* Prove: a − b = ( a − b ) � a .b
n n
� �
� �
�0 �
i=
n −1
�a i .bn −1−i
Assume a n − bn = is right
i =0

( )
a n +1 − b n +1 = a n +1
− a b + a n b − b n +1 = a n ( a − b ) + b a n − b n
n


�−1 i n −1−i � �−1 i n −1−i �
�n �
n n
= a ( a − b ) + b ( a − b ) � a .b � ( a − b ) . � + b. � a .b
n
� �
= �
a �
� � � �
� �
�0 � �0 �
i= i=
� �
�n n −1 i n −i � �n i n −i � n +1 �n i n −i �
n +1
= ( a − b ) . � + � .b � ( a − b ) � a .b � a = ( a − b ) � a .b �
� �
= −b

a a
� � � � � �
� � �0 � �0 �
i =0 i= i=
n −1
( ) ()
2� 2 n −1−i �
2n 2n 2 2i

�a −b = a −b � a . b �
� �
�0 �
i=




44
2n
2n +1 2n −i
Suppose a 2n +1 − ( −b ) �a i .b2n −i . ( −1)
= is right
i =0
a 2n +3 + b 2n +3 = a 2n +3 − a 2n +1 2
b + a 2n +1b 2 + b 2n +3

)
(
( ) 2n +1
= a 2n +1 a 2 − b2 + b2 a 2n +1 − ( − b )
2n
� i 2n −i �
( a − b ) ( a + b ) + b ( a − b ) � a .b .( −1) 2n −i �
2n +1 2

=a
� �
�0 �
i=

�2n + 2 �
2n
2 � i 2n −i �
( −1) 2n −i �
2n +1
= ( a − b ) .� �
+a .b + b . � a .b �
a
� �
� �
�0 �
i=
� �
2n
�2n + 2 2n + 2−i �
2n +1
.b + � i .b2n + 2−i ( −1)
= ( a − b ) .� +a
a a �
� �
� �
i =0
� +2
2n
2n + 2−i �
� a 2n +3 + b 2n +3 = ( a − b ) . �� a i .b 2n + 2−i ( −1) �
� �
�i =0 �


x 4 .dx
* I=+
( )
2
x2 + a2

x 4 .dx
( ) x
put x = a.tan t � dx = a tan 2 t + 1 .dt, t = arctan
* I=�
( )
2 a
x2 + a2

( ) ( tan 4 t + 1 − 1) .dt
a 5 .tan 4 t. tan 2 t + 1 .dt tan 4 t.dt
�I=� = a� = a�
( tan t + 1)
2
tan 2 t + 1 tan 2 t + 1
4 2
a

( )( )+
� tan 2 t − 1 tan 2 t + 1 dt �
)
(�
( tan )
dt
= a� � 2
t − 1 dt − � 2 t.dt
=
� � 2 t +1� a cos
� 2
tan t + 1 tan
� �
� �




45
( )
cos 2t + 1 � � dt 1 sin 2t �

= a � tan 2 t + 1 − 2 dt − � .dt � a � t − 2t − � − .
= tan
� �
2 22 2�
� ��
x 2 .dx
5at a sin 2t
= a tan t − − �I=�
( )
2
2 4
x2 + a2

x � 5a xa x�
� �
= a.tan �
arctan � .arctan − .sin �
− 2arctan �
a� 2 a4 a�
� �
a 2 .x
x 5a xa x� 5a x

= a. − .arctan − .sin � arctan � x − .arctan −
=
2
( )
a 4 a2 + x2
a2 a4 a� 2


x
x� ax
� a
= =
sin �2arctan �
2 2 2
a�
� � � a +x
x
1+ � �
a
��


dx
* I=+
( )
3
x2 + a2

dx a.dt x
* I=� put x = a.tan t � dx = , t = arctan
cos 2 t
( )
a
3
x2 + a2
1
a.dt a.dt 1
dt
�1 = a 3 � t.dt
�I=� =� = cos
3 3 2
( ) �1�a
� �
cos 2 t � a 2 tan 2 t + 1 � a 3 cos2 t � cos t

cos 2 t �
� � �
x
sin t 1 x� 1 x x
� a
= = = = =
sin �arctan �
a2 a2 a � a2
� 2 2 2 2
a2 a2 + x2
� � a a +x
x
1+ � �
a
��




46
dx
* I=�
5
� x2 + a2 �
� �
� �
dx a.dt x
* I=� put x = a.tan t � dx = , t = arctan
5
cos 2 t a
� x2 + a2 �
� �
� �
3
a.dt a.dt 1
dt
cos3
� 1 = a 5 � t.dt
�I=� =� =
5 5
a4
( )
2�2 � 2� 1 �
2 5
cos t � a tan t + 1 � a cos t � cos3 t
2�
� � � cos t �
�� �
x�
sin 3arctan �
�� �
1 cos 3t − 3cos t 1� sin3t � 1�� x�

a�
= .dt = − 3sin t �= − 3sin � �
4�
arctan �
4�3 4�
4 3 a�

4a � � 4a �
a
� �
� �


dx
* I=−
( )
3
a2 − x2

dx x
* I=� put x = a.sin t � dx = a.cos t.dt, t = arcsin
( )
a
3
a2 − x2

a.cos t.dt a.cos t.dt 1 dt tan t 1 x�

�I=� =� = = = .tan �
� arcsin �
( a ( 1 − sin t ) ) a 3 cos3 t a 2 cos 2 t a2 a2 a�

3
2 2

x
1 1 x x
a
= = =
a2 a2
2
a2 − x2 a2 a2 − x2
x
��
1− � �
a
��




47
dx
* I=−
( )
2n +1
a2 − x2

dx x
* I=+ put x = a.sin t � dx = a.cos t.dt, t = arcsin
( )
a
2n +1
a2 − x2


( )
n
a.cos t.dt a.cos t.dt 1 dt 1
tan 2
�I=� =� = � 2n t = a 2n � t + 1 .dt
a 2n +1 cos 2n +1 t a 2n cos
( a ( 1 − sin t ) )
2n +1
2 2


n −1 C k ( tan t ) 2k +1
n −1
( tan )
n −1
1 1 1
.d ( tan t ) = Cn tan 2k .d ( tan t ) = + n
2 k

= t +1
� � 2k + 1
2n
a 2n k =0 2n
a a k =0
2k +1
�x �
Ck �
x�
2k +1 � �
k n�
C tan arcsin �
� �
1 n −1 n a � 1 n −1 2 2
�a −x
� �
� �
= =
2k + 1 2k + 1
a 2n k =0 a 2n k =0


dx x
* I=� put x = a.sin t � dx = a.cos t.dt, t = arcsin
( a2 − x2 )
a
5



( )
2
a.cos t.dt a.cos t.dt 1 dt 1
tan 2
�I=� =� = � 4 t = a4 � t + 1 .dt
( a ( 1 − sin t ) ) a 5 cos5 t a 4 cos
5
2 2


1 � 3t
( )
� 1 � 3�
1 tan 1 x� x�
a

tan t + 1 .d ( tan t ) =
2
= + tan t �= +
� � � tan �
arcsin � tan �
arcsin �

a4 � 3 � a4 �
a4 3 a� a�
� � �
� �
3
1� x � x
= �+

4� 2 �42
3a � a − x 2 2
� a a −x




48
−a.cos t.dt
dx a a
* I== put x = � dx = , t = arcsin
sin 2 t
( )
sin t x
3
x2 − a2

−a.cos t.dt −a.cos t.dt −1 cos t.dt −1 sin t.dt
�I=� =� = =
� �
3 2
cos3 t a 2 cos 2 t
3 2 cos t a
3
�2 � 1 �
� a sin t.
sin 2 t � � − 1�
a � sin 3 t sin t
sin 2
�� t � �
1 d ( cos t ) −1 −1 −1
= � 2 t = a 2 .cos t = 2 � a � =
a 2 cos a�
a .cos � csin � a 2 . 1 − sin 2 �
ar arcsin �

x�
� x�

−1 −x
= =
a2 a2 x2 − a2
2
1−
a
x2


dx
* I=−
( )
2n +1
x2 − a2

−a.cos t.dt
dx a a
* I=� put x = � dx = , t = arcsin
sin 2 t
( )
sin t x
2n +1
x2 − a2

−a.cos t.dt
�I=�
2n +1
� �1 �

sin t �2 �
2
− 1�
a �
sin 2
�� t � �
( sin t ) 2n −1 .dt
−a.cos t.dt −1 −1
cos t.dt
=� = =
� �
2n +1
( cos t ) 2n +1 a 2n ( cos t ) 2n +1
a 2n
2 � t�
cos
2n +1
a sin t. � �
( sin t ) 2n −1
� t�
sin




49
( )
n −1
d ( cos t )
1 − cos 2 t
2n − 2
( sin t ) d ( cos t )
1 1
= =
� �
a 2n cos 2 t a 2n cos 2 t
n −1
k
�Ck ( −1) cos2k t.d ( cos t )
n
1 k =0
= �
a 2n cos 2 t
n −1 C k ( −1) k ( cos t ) 2k −1
n −1
1 1
2k − 2
k
Ck ( −1) ( cos t ) .d ( cos t ) = n
� �
= � n
2k − 1
a 2n k =0 2n
a k =0

2k −1
2k −1
k� �
a�

k� � �
a� Ck ( −1) � 1 − sin 2 �
n −1 Cn ( −1) � �
k arcsin ��
cos arcsin � n
� � �
1 n −1 x�

x�
1 �� � � �
� �
= =
2k − 1 2k − 1
a 2n k =0 a 2n k =0
2k −1
� 2�
a
��
k
Ck ( −1) � 1 − � � �
n
� � ��x
1 n −1 � �

=
2k − 1
2n
a k =0



x 2 .dx
* I=− , a >0
2 2
x −a
x 2 .dx
1 2x.dx
* I=� , a >0 I= � put u = x � du = dx
x.
2
x2 − a2 x2 − a2
1
( ) =( )
2 − 2 +1
2 2 2 1
d x −a x −a
( )
2x.dx 2x.dx
= 2 x2 − a2 2
dv = �v=� =�
1
2 2
2 2
x2 − a2
x −a x −a − +1
2
x2 − a2
2 2 2 2 2 2
� I = u.v − � = x. x − a − � − a .dx = x. x − a − �
v.du x .dx
2 2
x −a
x 2 .dx
dx dx
2 2
+ a2 � = x. x 2 − a 2 − I + a 2 �
I = x. x − a − �
x2 − a2 x2 − a2 x2 − a2

� 2I = x. x 2 − a 2 + a 2 ln x + x 2 − a 2



50
* I = = ( x − a ) ( b − x ) .dx

* I = � x − a ) ( b − x ) .dx
(
dat x = a + ( b − a ) sin 2 t � dx = 2 ( b − a ) sin t.cos t.dt
( b − a ) ( 1 − cos 2t ) 2( x − a)
� ( 1 − cos 2t ) =
�x −a =
( b − a)
2
2( x − a) b − 2x + a � − 2x + a �
1 b
� cos 2t = 1 − = � t = .arccos � �
( b − a) b−a � b−a �
2


( ( b − a ) sin 2 t ) ( b − a − ( b − a ) .sin 2 t ) .2 ( b − a ) sin t.cos t.dt
�I==

= 2= ( ( b − a ) sin 2 t ) ( ( b − a ) cos 2 t ) . ( b − a ) sin t.cos t.dt

( b − a) 2
2
2 � 2t �
sin
2
= 2( b − a) t.cos t.dt = 2 ( b − a )
2 2 2
�.dt =
� �2 �
sin sin 2t.dt

2
� �
( b − a)2 2 2
( b − a ) t ( b − a ) sin 4t
1 − cos 4t
= .dt = −
�2
2 4 16

b
* I = � x − a ) ( b − x ) .dx
(
a
put x = a + ( b − a ) sin 2 t � dx = 2 ( b − a ) sin t.cos t.dt
π
x = a � ( b − a ) sin 2 t = 0 � sin t = 0 � t = 0 x = b � sin 2 t = 1 � t = �
2
π
π
2 2
( b − a) 2 π
�b − a ) t ( b −− ) sin 4t �2
( a
�I=� �=

�4 16 8

� �
0




51
dx
* I=−
( x − a) ( b − x)
dx
* I=−
( x − a) ( b − x)
put x = a + ( b − a ) sin 2 t � dx = 2 ( b − a ) sin t.cos t.dt
( b − a ) ( 1 − cos 2t ) 2( x − a)
� ( 1 − cos 2t ) =
�x−a =
( b − a)
2
2( x − a) b − 2x + a � − 2x + a �
1 b
� cos 2t = 1 − = � t = .arccos � �
( b − a) b−a � b−a �
2
2 ( b − a ) sin t.cos t.dt 2 ( b − a ) sin t.cos t.dt
�I=− =�
( ( b − a ) sin 2 t ) ( b − a − ( b − a ) .sin 2 t ) ( ( b − a ) sin 2 t ) ( ( b − a ) cos2 t )
� − 2x + a �
b
= 2� = 2t = arccos � +
dt �C
� b−a �




dx
* I=−
( ( x − a) ( b − x) )
3


dx
put x = a + ( b − a ) sin 2 t � dx = 2 ( b − a ) sin t.cos t.dt
* I=�
( ( x − a) ( b − x) )
3


( b − a ) ( 1 − cos 2t ) 2( x − a) 2( x − a) b − 2x + a
� ( 1 − cos 2t ) =
�x−a = � cos 2t = 1 − =
( b − a) ( b − a) b−a
2
� − 2x + a �
1 b
� t = .arccos � �
� b−a �
2
2 ( b − a ) sin t.cos t.dt 2 ( b − a ) sin t.cos t.dt
�I=� =�
3 3
( )( ) ( )( )
� � � �
� ( b − a ) sin t b − a − ( b − a ) .sin t � ( b − a ) sin t ( b − a ) cos t
2 2 2 2
� �
� � � �




52
2 dt 2 dt 8 dt
= = =
2�2 2� 2�2
2 2
( b − a) ( b − a)
� 2t � ( b − a )
sin t.cos t sin 2t
sin
� �
�2 �
− cot 2t −1 −1
dt 1
� 2 2t = 2 = − 2.tan 2t = =
� − 2x + a �
� �
b 2
sin
� 2 1 − � − 2x + a �
b
2.tan �arccos � �
� b−a � � �
� � � b−a �
� − 2x + a �
b
� �
� b−a �

� − 2x + a � � − 2x + a �
b b
� � � �
−4
� b−a � � � b−a �
1 8 dt
=− . =
2�2
.
sin 2t ( b − a ) 2
( b − a)
2 2 2
� − 2x + a � � − 2x + a �
b b
1− � 1− �
� �
� b−a � � b−a �
−4 b − 2x + a
= .
( b − a)3 2
� − 2x + a �
b
1− � �
� b−a �


'
� �
� � − 2x + a � �
b
� � b−a � �
−4
.� � �
Check the solution : I = �
( b − a ) 2 � � − 2x + a � �
2
b
� 1− � ��
� � b−a ��
� − 2x + a � ' −2
b
put y = � � y=

� b−a � b−a




53
'
2� 2�
' '
'
� y 1 − y − �1 − y �.y
� � �
−4
�y y� � �
� �
�I= =
,
( b − a ) 2 � 1 − y2 1 − y2
� � 1 − y2 �
� � � �

( )
y' 1 − y 2 + y 2 .y'
−2y.y'
y' 1 − y 2 − .y
2 1 − y2 1 − y2
= =
( 1 − y2 )
1 − y2

−2
'
y b−a
= =
3
(1− y ) 23 �2 �
( x − a) ( b − x) �

�b − a )
( �


( b − a ) 2 − ( b − 2x + a ) 2
2
(1− y ) � − 2x + a �
b
2
= 1− � �=
� b−a � ( b − a) 2
( )=
b 2 + a 2 − 2ab − b2 + 4x 2 + a 2 − 4ax − 4bx + 2ab 4ax + 4bx − 4ab − 4x 2
=
( b − a) 2 ( b − a) 2
4a ( x − b ) + 4x ( b − x ) 4( x − a) ( b − x) 2
( x − a) ( b − x)
= = =
( b − a)
2 2
( b − a) ( b − a)

� �
� � − 2x + a � �
b
� � b−a � �
−4 8
.� � �=
�I= �
( b − a ) 2 � � − 2x + a � � 3
2
3� 2 �
b
� � ( b − a) � ( x − a) ( b − x)
�1− � �
( b − a)
b−a ��
�� � �
1
=
( ( x − a) ( b − x) )
3




54
dx
* I=−
( ( x − a) ( b − x) )
2n +1


dx
* I=�
( ( x − a) ( b − x) )
2n +1


put x = a + ( b − a ) sin 2 t � dx = 2 ( b − a ) sin t.cos t.dt
( b − a ) ( 1 − cos 2t ) 2( x − a)
� ( 1 − cos 2t ) =
�x −a =
( b − a)
2
2( x − a) b − 2x + a � − 2x + a �
1 b
� cos 2t = 1 − = � t = .arccos � �
( b − a) b−a � b−a �
2

2 ( b − a ) sin t.cos t.dt
�I=�
2n +1
( )( )
� �
� ( b − a ) sin t b − a − ( b − a ) .sin t
2 2

� �
2 ( b − a ) sin t.cos t.dt
=�
2n +1
( )( )
� 2�
� ( b − a ) sin t ( b − a ) cos t �
2
� �

22n +1
2 dt 2 dt dt
= = =
� � � 2n 2t
( b − a ) 2n ( sin t.cos t ) 2n ( b − a ) 2n 2n
( b − a ) 2n sin
� 2t �
sin
� �
�2 �
22n +1 22n +1
( ) ( )
n −1
n n −1
n
�1)
(− �1)
(− d ( cot 2t )
cot 2 2t + 1 cot 2 2t + 1
= dt =
2n 2n
( b − a) ( b − a)
n −1
( −1) n −1 n −1 Ck −1 ( cot 2t ) 2k +1
22n +1 ( −1) 2n +1
n −1 2
2k
Ck −1 ( cot 2t ) .d ( cot 2t ) = .+ n

= � n
2k + 1
( b − a ) 2n ( b − a ) 2n k =0
k =0




55
2k +1
� − 2x + a ��
k�� �
b
n −1 n −1 C n −1 � �
cot arccos � ��

22n +1 ( −1) � b − a ��
�� �

= .
2k + 1
( b − a ) 2n k =0
2k +1
� �
� − 2x + a � �
b

� �
k � � b−a � �
Cn −1 � �
2
� � − 2x + a � �
b
� 1− � b − a � �
n −1 n −1
2n +1
( −1)
2 �� ��
.−
= +C
2k + 1
( b − a ) 2n k =0


( ( x − a) ( b − x) )
3
* I== .dx

( )
3
* I = �( x − a) ( b − x) .dx

put x = a + ( b − a ) sin 2 t � dx = 2 ( b − a ) sin t.cos t.dt
( b − a ) ( 1 − cos 2t ) 2( x − a)
� ( 1 − cos 2t ) =
�x −a =
( b − a)
2
2( x − a) b − 2x + a � − 2x + a �
1 b
� cos 2t = 1 − = � t = .arccos � �
( b − a) b−a � b−a �
2

3
( )( )
� �
� I = � ( b − a ) sin 2 t b − a − ( b − a ) .sin 2 t �.2 ( b − a ) sin t.cos t.dt

� �
3
( )( )

( b − a ) cos t �. ( b − a ) sin t.cos t.dt
= 2� ( b − a ) sin 2 t 2
� �
� �
( b − a) 4
4
2 � 2t �
sin
4
= 2( b − a) t.cos t.dt = 2 ( b − a )
4 4 4
�.dt =
� �2 �
sin sin 2t.dt

8
� �
( b − a)4 ( b − a ) 4 � 8t
d ( 2t ) sin sin 4t 6t �
4
= = − +�

sin 2t. �
8 2 16 � 32 4 8�




56
b
( )
3
* I = �( x − a) ( b − x) put x = a + ( b − a ) sin 2 t � dx = 2 ( b − a ) sin t.cos t.dt
.dx
a
π
x = a � ( b − a ) sin 2 t = 0 � t = 0, x = b � sin 2 t = 1 � t = �
2
π
π
4 4 4
�b −6 ) sin 8t sin 4t �
( a� 3π ( b −4 ) 3π ( b − a )
6t � 2
a
�I=� �
− +�= =

27
� 16 � 32 4 8� 8.16

� �0

2
( )
� − cos 2x �
1 1
4
cos 2
* � x.dx = � �.dx = � 2x − 2.cos 2x + 1 dx
sin �
�2 4

x 1 1 + cos 4x
1 1 sin 2x x
= � 2 2xdx − � 2xdx + = � dx − +
cos cos
4 2 44 2 4 4
1 � sin 4x � sin 2x x sin 4x sin 2x 3x
= �+ − += − + +C
x �
8� 4�4 4 32 4 8
2
( )
� + cos 2x �
1 1
4
cos 2
* � x.dx = � �.dx = � 2x + 2.cos 2x + 1 dx
cos �
�2 4

x 1 1 + cos 4x
1 1 sin 2x x
= � 2 2xdx + � 2xdx + = � dx + +
cos cos
4 2 44 2 4 4
1 � sin 4x � sin 2x x sin 4x sin 2x 3x
= �+ + += + + +C
x �
8� 4�4 4 32 4 8




57
dx
* I=+
( x + a ) ( x + b)
dx
( x + a ) ( x + b)
Put t = x + a + x + b � t 2 = 2x + a + b + 2
* I=+
( x + a ) ( x + b)
'
� + a ) ( x + b) �
(x
t 2 − ( 2x + a + b ) dx
, 2t.dt = 2dx + 2 � �
( x + a ) ( x + b) =

2 ( x + a ) ( x + b)
2

� + a + b + 2 ( x + a ) ( x + b) � t 2 .dx
2x t.dx
=� �= = 2t.dt � dt =
dx
( x + a ) ( x + b) ( x + a ) ( x + b) ( x + a ) ( x + b)
� � 2
� �
( x + a ) ( x + b ) .dt 2 ( x + a ) ( x + b ) .dt
2 dt
� dx = �I=� = 2� = 2ln t + C
t. ( x + a ) ( x + b )
t t
dx
�I=+ = 2 ln x+a + x+b +C
( x + a ) ( x + b)

dx
* I=+
x +1 + x −1 + 2
dx
* I=�
x +1 + x −1 + 2
x +1− x +1 2
Put u = x + 1 + x − 1 = = >0
x +1 − x −1 x +1 − x −1
2 2
� x + 1 − x − 1 = � 2 x + 1 = u + � 4 ( x + 1)
u u
� 8�
4
= u2 + 4 + � 4dx = � −
2u du
3�
2
� u�
u

8 4 4
� �
2u − u− � u4 �
4
1 u −4
2 2 2
1 1
I = � u du = � u du = � du − � u
du == �� du �
4 ( u + 2) 2 u+2 2 u 4 + 2u 3 2 � u 4 + 2u 3 u 2 + 2u �
� �
� �
4
2 2 2.dk 2
I1 = − � u du dat u = � du = − ,k=
u 2 + 2u k2
k u



58
k2 k2 u2 u 2 + 2u − 2u + 2 − 2
� 2.dk � 1
� I1 = − � − =� I2 = =
dk
4 � k2 � 2 ( 1 + k )
4 u 2 + 2u u 2 + 2u
+� �
2k
k
− ( 2u + 2 ) 2
=1+ +
u 2 + 2u u 2 + 2u
1 � u2 k2 �
1
�I= � du + �
� dk �
2 (1+ k) �
2 � u 2 + 2u

d(1+ k) �
� 1 �k 2 − 1
2u + 2
1� 2
= �du − � du + � +� dk + �
� u 2 + 2u du � � �
u 2 + 2u � 4 � 1 + k 1+ k �
2� � �
� �
1� �1
1
du � �( k − 1) dk + ln 1 + k �
= � − ln u 2 + 2u + 2.� +�
u
� 2�2 � 4� �
2�
1+ �.u

� �
� u�
� �

2�

d � + 1�
−2du
2 1 2
u�

I3 = � du = − � = −� = − ln 1 +
� 2 � u2
� 2�2 � 2� u
1+ � �+ � �+ �
.u 1 1

� u� � u� � u�
2 � 1 �2 �
1� k
2
� I = � − ln u + 2u − ln 1 + � � − k + ln 1 + k � C
+ +
u
2� u � 4� 2 �
� �
2
where 0 < u = = x + 1 + x − 1
k


x
* I=− dx
3
(1− x)
t2
x�1 � x x 2t.dt
* I=� dx = � put t = �x= � dx =
dx
� �
�− x � ( 1 − x ) (1− x)
(1+ t )
3
1 + t2 22
( 1− x) 1

t 2 .dt
( ) � 1�
2t.dt
2
I = �+ t = 2� = 2� − � = 2t − 2arctan t
1 t. 1 dt
( )

( )
22 � 1 + t2 �
2
1+ t
1+ t

x x x
�I=� dx = 2 − 2arctan
(1− x) (1− x)
(1− x)3
59
60
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