Các dạng bài tập tích phân

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  1. Chapter 5e Integral of Irrational Function 8/ ..................................................................................................................................2 ................................................................................................................................................5 * ..........................................................................................................................................6 ................................................................................................................................................7 ................................................................................................................................................9 ..............................................................................................................................................11 ..............................................................................................................................................12 * ..........................................................................................................................................13 * ..........................................................................................................................................14 * ..........................................................................................................................15 ..............................................................................................................................................17 * ..........................................................................................................................................18 * ...........................................................................................................................................19 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...........................................................................................................................................29 .............................................................................................................................................31 ..............................................................................................................................................32 .............................................................................................................................................33 ..............................................................................................................................................36 ..............................................................................................................................................37 ..............................................................................................................................................37 ..............................................................................................................................................38 ..............................................................................................................................................38 ..............................................................................................................................................39 ..............................................................................................................................................39 ..............................................................................................................................................39 ..............................................................................................................................................40 ..............................................................................................................................................41 ..............................................................................................................................................42 ..............................................................................................................................................42 ..............................................................................................................................................43 ..............................................................................................................................................44 * ...........................................................................................................................................44 ..............................................................................................................................................45 ..............................................................................................................................................46 ..............................................................................................................................................47 ..............................................................................................................................................47 1
  2. ..............................................................................................................................................48 ..............................................................................................................................................49 ..............................................................................................................................................50 ..............................................................................................................................................51 ..............................................................................................................................................52 ..............................................................................................................................................52 ..............................................................................................................................................55 ..............................................................................................................................................56 ..............................................................................................................................................58 ...............................................................................................................................................58 ..............................................................................................................................................59 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
  3. ( ) 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
  4. 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
  5. 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
  6. 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
  7. � � � 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
  8. 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
  9. +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. −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
  11. 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
  12. 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 ) , xi [ 1, c] , * Put f ( x i ) = xi m . a + b.xi n1 i ( c − 1) ) ( n np c −1 � I = lim �m h m +1. a + b. ( i.h ) 1 = ∆x = , xi = i n n n + +m i =1 h = +m p m +1 � n1 � m �c − 1) � ( �( c − 1) � n c −1 i . � + b. � � � Dat h = � = lim i a � � � �� �n �n n n + +m � i =1 � � c= +m p m� n1 � � − 1 � �( c − 1) � �( c − 1) � ni i c + � . � + b. � �� = lim � � � a � �� n = +m � n � 1� n �n �� i= � c= +m Ci .bi a p−i c p � ( xi ) .∆x = c lim. � .( a + b.x ) n np p m .dx = −� = lim f x i =1 ( n1.i + m + 1) n − +i i =1 + m 1 c= +i We have the formula: Pn ( x ) dx dx = Q n −1 ( x ) ax 2 + bx + c + p.� ( 1) �2 2 ax + bx + c ax + bx + c Pn ( x ) is a n degree polynomial, Qn −1 ( x ) is a n – 1 degree polynomial with indefinite coefficients To determine p and coefficients of Qn −1 ( x ) , we take derivative (1) and equate coefficients of two sides to obtain system of equations. (How to prove this formula?) 12
  13. x3 − x + 1 ( ) 1 5 x 2 + 2x + 2 + .ln � + 1 + x 2 + 2x + 2 � 2x 2 − 5x + 1 * I=+ dx = x � � 2� � 6 x 2 + 2x + 2 x3 − x + 1 ( ) dx dx = ax 2 + bx + c x 2 + 2x + 2 + d.� VD : I = � x 2 + 2x + 2 x 2 + 2x + 2 Take derivative in two side, we obtain : x3 − x + 1 ( )2 2x + 2 d = ( 2ax + b ) x 2 + 2x + 2 + ax 2 + bx + c + x 2 + 2x + 2 x 2 + 2x + 2 x 2 + 2x + 2 ( )( ) � x 3 − x + 1 = ( 2ax + b ) x 2 + 2x + 2 + ax 2 + bx + c ( x + 1) + d = 2ax 3 + 4ax 2 + 4ax + bx 2 + 2bx + 2b + ax 3 + bx 2 + cx + ax 2 + bx + c + d = 3ax 3 + x 2 ( 5a + 2b ) + x ( 4a + 3b + c ) + ( 2b + c + d ) � 3a = 1, 5a + 2b = 0, 4a + 3b + c = −1, 2b + c + d = 1 x3 − x + 1 1 5 1 5 �a = ,b=− ,c= ,d = �I= + dx 3 6 6 2 2 x + 2x + 2 ( ) 1 5 dx = 2x 2 − 5x + 1 x 2 + 2x + 2 + .+ 6 2 x 2 + 2x + 2 d ( x + 1) dx = ln � + 1 + x 2 + 2x + 2 � =� �2 x � � � � 2 ( x + 1) + 1 x + 2x + 2 � dx � � + x2 + a � = ln � � � � x � � � � x2 + a � � � x3 − x + 1 ( ) 1 5 x 2 + 2x + 2 + .ln −x + 1 + x 2 + 2x + 2 � � dx = 2x 2 − 5x + 1 �I=� � 2� � 6 x 2 + 2x + 2 13
  14. 1 dx 1 * I=+ = ( ) n2 nn 1 + xn 0 1+ x 1+ n 1 1 ( ) n−n dx I=� =� 1+ x dx ( ) nn n 0 1+ x 1+ x 0 zn 1 1 −n n n +1 = z �x= � 1+ x =1+ = Put: x 1 n zn − 1 z −1 ( ) n z −1 n z n −1dz z n −1dz − n −1 n −1 n +1 n −1 � −n.x dx = n.z dz � dx = − = −x dz = − .z x − n −1 n +1 ( ) n z −1 n −x = 1 � z = n 2 � � 1 change the limits � � = 0 � z = lim �+ 1 � = +� n x 1 � n� � x + 0+ � x � � 1 1 1 ( ) n −1− n dx So I = � = �+ x 1 dx ( ) nn n 0 1+ x 1+ x 0 1+ n − n2 z n −1dz � zn �n = − lim �z n − 1 � � . � � n +1 a + +1 ( ) a� � zn − 1 n n2 n2 n z −1−n .z n −1dz 12 −2 = − lim = − lim � dz = lim � z 1+ n n +1 a +z z a + +− a +m z z ( ) ( ) − a a a z − 1 n . zn − 1 n n 1 1 1 = − = n2 a n2 14
  15. dx * I=+ x +1 + x −1 + 2 dx 15/ 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 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
  16. 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
  17. 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
  18. 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
  19. 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
  20. 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
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