Bài 7. Tích phân hàm vô t
199
BÀI 7. TÍCH PHÂN HÀM VÔ T
I. TÍCH PHÂN CÓ CHA CÁC CN THC CA BIN C LP

Xét dng c bn thng gp:
( )
p
m n
I x a bx dx
= +
vi
m, n, p
hu t
1.1.
Nu
p
∈
Z
thì gi
k
là mu s chung nh nht ca các phân s ti gin biu
th bi
m
và
n
, khi ó t
x
=
t
k
.
1.2.
Nu
1
m
n
+
∈
Z thì gi
S
là mu s ca
p
và t
n s
a bx t
+ =
1.3.
Nu
1m
n
+
+ ∈
thì gi
S
bng mu s ca
p
và t
n
s
n
a bx
t
x
+
=

Xét
1
1
j
j
r
r
q
q
I R x, x ,..., x dx
=
vi
r
1
, q
1
,…r
j
, q
j
là các s nguyên dng.
Gi
k
là bi s chung nh nht ca các mu s
q
1
, …, q
j
. Khi ó ta có:
1 1
1
j j
j
r
r;...;
q k q k
= =
α
α
. t
(
)
( )
11
1
j
k k k
x t I R t ,t ,....,t kt dt R t a
−
== =
α
α

Xét
()()
m r
n s
ax b ax b
I R x, ,..., dx
cx d cx d
+ +
=+ +
vi
m, n, …, r, s
nguyên dng
t
ax b
t
cx d
+
=
+
( )
2
t d b ad bc
x ;dx dt
a ct a ct
− −
= =
−−
( )
2
m r
n s
td b ad bc
I R ,t ,...,t dt
a ct a ct
− −
=
−
−
Gi
k
là bi s chung nh nht ca các s:
{
}
n,...s
. t
t
=
u
k
thì
( )
( )
1 1 1
2 2
k
mrm r k
n s
k
td b ad bc u td b ad bc
I R ,t ,...,t dt R ,u ,...,u ku d u
a ct a ct
a ct a cu
−
− − − −
= =
− −
−−
II. CÁC BÀI TP MU MINH HA
1. Dng 1:
( )
p
m n
I = x a + bx dx
vi m, n, p
∈
∈∈
∈
Q
•
(
)
1
1 3
4 4
1
x x dx
−
= +
4
143
xdx
I =
1 + x
1 3
m ; n ; p 1 Z k 4
4 4
−
= = = − ∈ =
Chơng II: Nguyên hàm và tích phân
−
−−
−
Trn Phơng
200
t
4
t x
=
4 3
4
x t dx t dt
==
4
4
13 3
3
4
x dx 4t dt 4t
I 4t dt
1 t 1 t
1 x
= = = −
+ +
+
( ) ( )
( )
( )
2
2
t 1 t 1
2t 2 dt
t 1 t t 1
+ + −
= −
+ − +
(
)
( )
( )
2 2
2
22
dt t t t 1
2t 2 2 dt
t t 1
t 1 t t 1
− − +
= − −
− + + − +
()
2
2
2 3
2
dt t dt dt
2t 2 2 2
t 1
t 1
3
1
t2 2
= − − +
+
+
− +
2 3
4 2t 1 2
2t arctg ln 1 t 2 ln 1 t c
3
3 3
−
= − − + + + +
43
44
4 2. x 1 2
2 x arctg ln 1 x 2ln 1 x c
3
3 3
−
= − − + + + +
•
( )
( )
2
1 2 1 3
1
x x dx
−
= +
22
3
xdx
I =
1 + x
1 1
m ; n ; p 2 Z
2 3
= = = − ∈
k
=
6
t
6 5
66
t x x t dx t dt
===
. Khi ó:
(
)
( ) ( )
3 5 8
2
2 2
2 2
t 6t dt 6t dt
I
1 t 1 t
= =
+ +
( ) ( )
2
4 2 4 2
2 2 2
2 2
4t 3 4 dt
6 t 2t 3 dt 6 t 2t 3 dt 6
t 1
t 1 t 1
+
= − + − = − + − +
+
+ +
=
5 3
t 2t
6 3t 4 arctg t 6J
5 3
− + − +
. t
( )
2 2
2
dt dt
I ; J
t 1
t 1
= =
++
Ta có:
( )
2
2 2 2 2 2
2
1 t t t dt
1
I dt td 2
t 1 t 1 t 1 t 1
t 1
= = − = +
+ + + +
+
(
)
( ) ( )
2
2 2 2 2 2 2
2 2
t t 1 1 t dt dt t
2 dt 2 2 2I 2J
t 1 t 1 t 1 t 1
t 1 t 1
+ −
= + = + − = + −
+ + + +
+ +
( )
22
t t 1
2J I J arctg t c
2
t 1 2 t 1
= + = + +
++
( )
5 3
22
t 2t t 1
I 6 3t 4 arctg t 6 arctg t c
5 3 2
2 t 1
= − + − + + +
+
5 3
2
5 38 8 8
8
4
3 6t 20t 90t
21arctg t c
5
t 1
6 x 20 x 90 x
3
21arctg x c
5
x 1
− +
= − + +
+
− +
= − + +
+
Bài 7. Tích phân hàm vô t
201
•
( )
1 2
2 3
1
x x dx
= +
332
xdx
I =
1 + x
2 1 1
1; ; 3
3 2
m
m n p
n
+
= = = = ∈
t
( ) ( )
3 2
3 3
2 2 2 2 2 2
1 1 1 2 6 1
t x t x t x x dx t t dt
= + = + − = = −
( ) ( ) ( )
2
22
2 4 2
3
23
x dx 3t t 1 dt
I 3 t 1 dt 3 t 2t 1 dt
t
1 x
−
= = = − = − +
+
(
)
(
)
(
)
5 2 3 2
5 3 2 2 23 3 3
3 3
t 2t 3t c 1 x 2 1 x 3 1 x c
5 5
= − + + = + − + + + +
•
( )
1 3
3 3
2
x x dx
−
−
= −
−
43
3 3
dx
I =
x 2 x
1 1
3; 3, 1
3
m
m n p p
n
− +
= − = = + = − ∈
t
3
3
2
x
t
x
−
=
( )
3 2
3 3 2
3 3 3 2
3
2 2 2 2
11
1
x t dt
t x x dx
x x t t
− −
= = − ==
++
( )
2 2
4
2 2
3 3 3
3 3 3
6
3
dx x dx 1 2t dt
I
2
x 2 x 2 x
t 1
t
xt 1
x
−
= = = ⋅
− − +
+
2
3
2 3
1 t 2 x
t dt c c
2 4 2x
− − −
= = + = − +
•
−
33
5
I = 3x x dx
1 1 1
; 2, 1
3 3
m
m n p p
n
+
= = = + = ∈
t
( )
33 3 2
3 2
3 2 3 2
3
3 3 3 3 9
1 2
1
1
x x x x t dt
t t x x dx
xx x t t
− − −
== = − ==
++
( )
( )
3 3
3
3
3
52 3
3
1 3x x 9 t dt 3 1
I 3x x dx 2x dx td
2 x 2 2
t 1
t 1
− −
= − = = =
+
+
( ) ( )
3
3 3
3t 3 dt 3t 3
I
2 2
t 1
2 t 1 2 t 1
= − = −
+
+ +
vi
3
1
dt
It
=
+
( )
( )
2
1 1
dt
I
t t t
=
+ − +
(
)
( ) ( ) ( )
( )
2 2
d t 1 du
u u 3u 3
t 1 t 1 3 t 1 3
+
= =
− +
+ + − + +
(
)
(
)
( )
( )
2 2
2
2
1 u 3u 3 u 3u 1 du u 3 du
du
3 3 u
u 3u 3
u u 3u 3
− + − − −
= = −
− +
− +
( )
2 2
1 du 1 2u 3 du 3 du
3 u 2 2
u 3u 3 u 3u 3
−
= − +
− + − +
Chơng II: Nguyên hàm và tích phân
−
−−
−
Trn Phơng
202
(
)
()
2
2 2
2
2
1 du 1 d u 3u 3 3 du
3 u 2 2
3
u 3u 3 3
u2
4
1 1 u 2u 3
ln 3arctg c
3 2 u 3u 3 3
− +
= − +
− +
− +
−
= + +
− +
( )
( ) ( )
( )
2
2
1 t 1 1 2 t 1 3
ln arctg c
63 3
t 1 3 t 1 3
+ + −
= + +
+ − + +
2
2
1 t 2t 1 1 2t 1
ln arctg c
62 3 3t t 1
+ + −
= + +
− +
( )
2
52
3
3t 3 1 t 2t 1 1 2t 1
I ln arctg
2 6
2 3 3
t t 1
2 t 1
+ + −
= − +
− +
+
( )
3
3 3 3
3 3 3
x 3x x 1 x 3x x x 3 2 3x x x
ln arctg c
2 4 3 4 x 3
− − + − −
= − − +
•
( )
1 2
4 2
1
x x dx
−
−
= +
64 2
dx
I =
x 1 + x
1 m 1
m 4, n 2, p p 2 Z
2 n
− +
= − = = + = − ∈
t
( )
2 2
2 2
2 2 2 2
2
1 1 1 1
11
1
x x t dt
t t x x dx
xx x t t
+ + −
== = + ==
−−
( )
( )
( )
3
2
2
62
4 2 2 2
6
dx x dx t 1 t dt
I t 1 dt
t
x 1 x 1 x t 1
x
x
− −
= = = ⋅ = − −
+ + −
( ) ( )
3
3 2 2 2 2
3 3
t 1 x 1 x 2x 1 1 x
t c c c
3 x
3x 3x
− − + + − +
= + + = + + = +
•
( )
( )
41
1 1 2
7
1
1
x x dx
−
−
= +
4
1
dx
I =
x 1 + x
1
m 1, n , p 1 Z
2
= − = = − ∈
t
22
t x t x dx t dt
===
( ) ( )
( )
( )
2 2 2
72
1 1 1
2t dt dt 1 t t
I 2 2 dt
t 1 t t 1 t
t 1 t
+ −
= = =
+ +
+
( )
( )
22
1
1
1 1 4
2 dt 2 ln t ln t 1 2 ln 2 ln 3 ln1 ln 2 2 ln
t t 1 3
= − = − + = − − + =
+
Bài 7. Tích phân hàm vô t
203
2. Dng 2:
j
1
j
1
r
r
q
q
I = R x, x , ..., x dx
•
( )
1 2
1 3
1
1
x
dx
x x−
− −
=
+
132
x 1
I = dx
x + x
. Gi
k
=
BSCNN
(2, 3)
=
6
t
6 5
66
t x x t dx t dt
===
(
)
3 4
5 2
16 4 2 2
t 1 6 t t t 1
I 6t dt dt 6 t 1 dt
t t t 1 t 1
− − −
= ⋅ = = − −
+ + +
3 2 6 3
2t 6t 3ln 1 t 6arctg t c 2 x 6 x 3ln 1 x arctg x c
= − − + + + = − − + + +
•
( ) ( )
1 4 1 8
4
1
− −
=
+
8
4
24
x x
I = dx
x 1 + x
x x
dx
x x
. Gi
k
=
BSCNN
(4, 8)
=
8
t
8 7
88
t x x t dx t dt
===
( )
2
7
22
8 2
t t t 1
I 8t dt 8 dt
t 1
t t 1
− −
= =
+
+
28
4
4 ln 1 t 8 arctg t c 4 ln 1 x 8arctg x c
= + − + = + − +
•
( )
( )
1 2
1 3
1 1
1 1
− − +
=
+ +
33
1 1 + x
I = dx
1 + 1 + x
x
dx
x
. Gi
k
=
BSCNN
(2, 3)
=
6
t
6 5
61 1 6
t x x t dx t dt
= + = − =
3
5 6 4 3 2
32 2
3
1 1 x 1 t t 1
I dx 6t dt 6 t t t t t 1 dt
1 1 x 1 t t 1
− + − −
= = ⋅ = − + + − − + +
+ + + +
75432
2
t t t t t 1
6 t ln t 1 arctg t c
7 5 4 3 2 2
= − − − + + − − + + +
( ) ( ) ( )
7 5 4
6 6 6
6 3 6
6 6 4
1 x 1 x 1 x 2 1 x
7 5 5
6 1 x 3ln 1 x 1 arctg 1 x c
−
= + + + + + − +
+ + + + + + + +
•
5
8
33
1
dx
I =
x 1 + x
. t
3 2
33
t x t x dx t dt
===