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30 BÀI TOÁN TÍCH PHÂN
Tìmnguyênhàm I =
6x 3+8x +1
(3x 2+4)
x 2+1dx
Bài 1
Li gii: Ta có
6x 3 +8x +1
3x 2 +4= 2x + 1
3x 2+4
=⇒ I =
2x + 1
3x 2+4
1 x 2+1
dx =
2x x 2 +1
dx +
1
(3x 2 +4)
x 2 +1dx
Tính I 1 =
2x x 2 +1
dx
Đt
x 2+1= t , x 2+1= t 2, 2t dt =2x dx =⇒ I 1 = 2
t dt
t = 2t =2
x 2 +1
Tính I 2 =
1
(3x 2+4)
x 2 +1. dx
Đt t =
x 2
+1
x , x t = x 2+1, x
2
t 2
= x 2
+1, x 2
=1
t 2 −1 , 3x 2
+4=4t 2
−1
t 2 −1
x dx =− t dt
(t 2−1)2, dx
x t =− t dt
(t 2−1)2x 2t ,
dx x 2+1
= dt
1− t 2
I 2 =
t 2−1
4t 2 −1
dt
1− t 2 =
dt
1−4t 2 = 1
2
1
2t +1− 1
2t −1
dt = 1
4ln
2t +1
2t −1= 1
4ln
2
x 2+1+x
2
x 2+1−x
Vy I = 2
x 2+1+ 1
4ln
2
x 2 +1+x
2
x 2 +1−x +C
Tìmnguyênhàm I =
cos2 x
sin x +
3cos x dx
Bài 2
Li gii: Dùng pp h s bt đnh cos2x = (a sin x +b cos x )(sin x +
3cos x )+c (sin2x +cos2x )
cos2 x =−1
4sin x +
3
4cos x
(sin x +
3cos x )+ 1
4= −1
4(sin x −
3cos x )(sin x +
3cos x )+ 1
4
I = −1
4 (sin x −
3cos x )(sin x +
3cos x )+ 1
4
sin x +
3cos x dx
=−1
4
(sin x −
3cos x ) dx + 1
4
1
sin x +
3cos x dx
=1
4(cos x +
3sin x )+ 1
4
1
sin x
+
3cos x
dx
Ta tính J = 1
4
dx
sin x +
3cos x = 1
8
dx
cos(x − π
6)= 1
8
cos(x −
π
6 )
1− sin2(x − π
6)
dx
Đt t = sin(x − π
6) =⇒ dt =cos(x − π
6) dx
=⇒ J = 1
8
dt
1− t 2 = 1
16
1
t +1− 1
t −1
dt = 1
16ln
t +1
t −1= 1
16ln
sin(x − π
6)+1
sin(x − π
6)−1
Vy I = 1
4(cos x +
3sin x )+ 1
16ln
sin(x − π
6)+1
sin(x − π
6)−1
+C
Tìmnguyênhàm I =
x 3 +x 2
4
4x +5dx
Bài 3
Li gii:
1
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I =
x 3+ x 2
4
4x +5dx =
x 4+ x 3
4
4x 5 +5x 4dx
= 1
20
4x
5+5x 4− 1
4 d(4x 5+5x
4)= 1
15
4
(4x 5 +5x 4)3+C
Tìmnguyênhàm I =
cos2x +
2cos
x + π
4
e
sin x +cos x +1 dx
Bài 4
Li gii: Tacó cos2x +
2cos
x + π
4
= (cos x − sin x )(sin x +cos x +1)
I =
(cos x − sin x )(sin x +cos x +1)e sin x +cos x +1 dx
=
(sin x +cos x +1)e sin x +cos x +1 d (sin x +cos x +1)
=
(sin x +cos x +1) d
e sin x +cos x +1
=(sin x +cos x +1)e
sin x +cos x +1 −
e sin x +cos x +1 d (sin x +cos x +1)
=(sin x
+cos x
+1)e
sin x +cos x +1
−e
sin x +cos x +1
+C
=(sin x +cos x )e sin x +cos x +1+C
Tìmnguyênhàm I =
3
3x −x 3 dx
Bài 5
Li gii:
Đt t =3
3x −x 3
x =⇒ x 2 = 3
t 3+1=⇒ 2x dx = −9t 2 dt
(t 3+1)2
I = 1
2
3
3x −x 3
x 2x dx = −9
2
t 3 dt
(t 3+1)2 = 3
2
t d
1
t 3+1
= 3t
2(t 3 +1)− 3
2
dt
t 3+1
Tính J = dt
t 3+1 = d(t
+1)
(t +1)[(t +1)2 −3(t +1)+3] =1
2 (ln3(1− t )−2ln3t + ln(1+ t ))
Vy I = 1
2x
3
3x − x 3− 3
4
ln 3
1−
3
3x − x 3
x
−2ln3
3
3x −x 3
x + ln
1+
3
3x − x 3
x
+C
Tìmnguyênhàm I =
1
x 4+4x 3+6x 2 +7x +4dx
Bài 6
Li gii: Tng các h s bc chn bng tng các h s bc l nên đa thc mu nhn x =−1 làm nghim
I = dx
(x
+1)[(x
+1)3
+3]= 1
3 (x +1)3 +3− (x +1)3
(x
+1)[(x
+1)3
+3]
dx = 1
3 dx
x
+1−
(x +1)2
(x
+1)3
+3
dx =1
3
ln |x +1|− 1
3
d((x +1)3)
(x +1)3 +3
= 1
3ln |x +1|− 1
9ln |(x +1)
3 +3|+C
Tínhtíchphân I =1
0
x ln
x +
1+ x 2
x +
1+ x 2
dx
Bài 7
Li gii:
Đt u = ln(x +
x 2+1), dv = x dx
x +
x 2+1= x (
x 2+1−x ) dx
Suy ra du =1
+
x
x 2
+1x +
x 2 +1
dx = dx x 2+1
, v = 12
(1+x 2)
1
2 d(1+ x 2)−x 2dx = 13
[(1+ x 2)3
2 − x 3]
I = 1
3[(1+ x
2)
32 − x
3]ln(x +
1+ x 2)
1
0− 1
3
1
0
[(1+x 2
)32 − x
3]
d x 1+ x 2
2
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Mà J =
[(1+x 2
)3
2 −x 3
] d x
1+ x 2=
d x
1+ x 2 −
x 3d x 1+x 2
= arctanx − 1
3(x
2 −2)
x 2+1
Nên I = 1
3[(1+x 2)
32 −x 3]ln(x +
1+ x 2)
1
0− 1
3arctanx
1
0+ 1
9(x 2−2)
x 2 +1
1
0
Vy I = 1
3(
8−1)ln(1+
2)− π
12+ 1
9(2+
2)
Tínhtíchphân I = 1
2
0 x ln
1
+x
1− x dx
Bài 8
Li gii:
Vi u = ln 1+x
1−x , dv = x dx nên du = 2
1−x 2 dx , v = 1
2x 2
I =1
2x
2ln
1+x
1−x
12
0− 1
2
0
x 2
1−x 2 dx = 1
8ln 3+
1
2
0
1− x 2 −1
1−x 2 dx
=1
8ln 3+ 1
2− 1
2
12
0
1
1+x + 1
1− x
dx = 1
8ln 3+ 1
2− 1
2ln
1+ x
1− x
1
2
0= 1
2− 3
8ln 3
Tínhtíchphân I =π
0 e −x
cos2x dx
Bài 9
Li gii:
I =
π
0
e −x
cos2x dx =−
π
0
cos2x d(e −x
)=−e −x
cos2x
π0−2
π
0
e −x
sin2x dx
= e −π+1+2
π
0
sin2x d(e −x
)= e −π+1+2e
−x sin2x
π0−4
π
0
e −x
cos2x dx = 1
5(e −π+1)
Tínhtíchphân I =
3
0
x 5 +2x 3 x 2 +1
dx
Bài 10
Li gii: I =
3
0
x (x 4 +2x 2) x 2 +1
dx =
3
0
(x 4+2x
2) d(
x 2 +1)
I = (x 4+2x
2)
x 2+1
3
0−
3
0
x 2+1 d(x
4 +2x 2
)
Tính J =
x 2 +1 d(x 4+2x
2)=
4x (x
2 +1)
x 2+1 dx = 4
x (x 2 +1)2
x 2+1
dx
= 4
(
x 2+1)4 d(
x 2+1)= 4
5(x
2 +1)2
x 2 +1
Nên I = (x 4+2x
2)
x 2+1
3
0− 4
5(x
2 +1)2
x 2 +1
3
0
Tínhtíchphân I =
e
1
1+ x 2 ln x
x + x 2 ln x dx
Bài 11
Li gii:
I =
e
1
1+ x 2 ln x
x +x 2 ln x dx =
e
1
1
x 2 + ln x
1
x + ln x
dx =
e
1
1
x + ln x
1
x + ln x
dx +
e
1
1
x 2 − 1
x
1
x + ln x
dx
=e
1
dx −e
1
d
1
x + ln x
1
x + ln x
= x e
1− ln
1
x + ln x
e
1= e −1− ln
1
e +1
3
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Tìmnguyênhàm I =
2(1+ ln x )+ x ln x (1+ ln x )
1+ x ln x dx
Bài 12
Li gii: Đt u = 1+x ln x =⇒ du = (1+ ln x ) dx
I = (2+ x ln x )(1+ ln x )
1
+x ln x
dx = u +1
u du =u + ln |u |+C =1+x ln x + ln |1+ x ln x |+C
Tínhtíchphân I = π
4
0
x 2(x 2 sin2x +1)− (x −1)sin2x
cos x (x 2 sin x +cos x )dx
Bài 13
Li gii:
I =
x 4 sin2x +x 2− (x −1)sin2x
x 2 sin x cos x +cos2 x dx =
π
4
0
2x 4 sin2x +2x 2−2x sin x +2sin2x
x 2 sin2x +cos2x +1dx
= π
4
0
2x 2(x 2 sin2x +cos2x +1)− (x 2 sin2x +cos2x +1)
x 2 sin2x +cos2x +1dx
=π
4
0
2x 2 dx −
π
4
0
d(x 2 sin2x +cos2x +1)
x 2 sin2x +
cos2x +
1
= 2
3x
3 π
4
0− ln |x
2sin2x +cos2x +1|
π
4
0= π
3
96+ ln 2− ln
π
2
16+1
Tìmnguyênhàm I =
(x 2 +1)+ (x 3 +x ln x +2)ln x
1+ x ln x dx
Bài 14
Li gii:
I =
(x 2+ ln x )+ x ln x (x 2+ ln x )+ (1+ ln x )
1+x ln x dx =
(x 2+ ln x )(1+ x ln x )+ (1+ ln x )
1+ x ln x dx
=(x 2
+ln x ) dx
+d(1+ x ln x )
1+
x ln x =
1
3.x
3
+x ln x
−x
+ln
|1
+x ln x
|+C
Tìmnguyênhàm I =
x 2(x 2 sin2 x +sin2x +cos x )+ sin x (2x −1− sin x )+1
x 2 sin x +cos x dx
Bài 15
Li gii: Vì x 2(x 2 sin2 x +sin2x +cos x )+ sin x (2x −1− sin x )+1= (x 2 sin x +cos x )2+ (x 2 sin x +cos x )
I =
(x 2
sin x +cos x ) dx +
d(x 2 sin x +cos x )
x 2 sin x +cos x =
x 2
sin x dx +sin x + ln |x 2
sin x +cos x |
Tính J =
x 2
sin x dx =−
x 2 d(cos x )=−x
2cos x +2
x cos x dx =−x
2cos x +2
x d(sin x )
J
=−x
2cos x
+2x sin x
−2sin x dx
=−x
2cos x
+2x sin x
+2cos x
Vy I =−x 2 cos x +2x sin x +2cos x + sin x + ln |x 2 sin x +cos x |+C
Tìm nguyên hàm I =
x (x +2)(3sin x −4sin3
x )+2cos x (cos x −2sin x )+3x 2
cos3x −1
e
x dx
Bài 16
Li gii: x (x +2)(3sin x −4sin
3x )+2cos x (cos x −2sin x )+3x
2cos3x −1
e
x
=
x 2
sin3x + (x 2
sin3x )+cos2x + (cos2x )
e x
=⇒ I = (x 2
sin3x +cos2x )e x
Tìmnguyênhàm I =
2x 4 ln2x + x ln x (x 3+1)+x − 1
x 2
1+ x 3 ln x dx
Bài 17
Li gii:
4
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2x 6ln2x + x 6 ln x +x 3 ln x +x 3−1
x 2+ x 5 ln x = 2[(x 3 ln x )
2 −1]+ x 3(x 3 ln x +1)+ (x 3 ln x +1)
x 2(1+ x 3 ln x )
= (x 3 ln x +1)(2x 3 ln x + x 3 −1)
x 2(1+ x 3 ln x )= 2x ln x +x − 1
x 2
Nên I =
2x ln x +x − 1
x 2
dx = 1
2x
2 + 1
x +
2x ln x dx = 1
2x
2 + 1
x +
ln x d(x 2
)
I = 1
2x
2 + 1
x +x
2ln x −x dx = 1
x + x
2ln x +C
Tìmnguyênhàm I =
x 2
sin(ln x ) dx
Bài 18
Li gii: Đt x = e t , ln x = t , dx = e t dt
=⇒ I =
e 3t
sin t dt =−e 3t
cos t +
3e 3t
cos t dt =−e 3t
cos t +3e 3t
sin t −
9e 3t
sin t dt
=⇒ 10I =3e 3t
sin t −e 3t
cos t =⇒ I = 1
10
3.e
3 ln x sin(ln x )−e
3 ln x cos(ln x )
+C
Tìmnguyênhàm I =
e x (x −1)+2x 3 +x 3(e x +x (x 2+1))
e x .x + x 2(x 2+1)dx
Bài 19
Li gii: e x (x −1)+2x 3 +x 3(e x +x (x 2+1))
e x .x + x 2(x 2+1)= x 3−1
x + 3x 2+e x +1
x 3+ x +e x = x 2− 1
x + (x 3+ x +e x )
x 3 +x +e x
Do đó
I = x 3
3− ln |x |+ ln |x 3+ x +e x |+C
Tínhtíchphân I
=π
3
π
6
ln(tan x ) dx
Bài 20
Li gii:
I = π
3
π
6
ln(tan x ) dx =đi bin (x = π
2−x )
π
3
π
6
ln(cot x ) dx =⇒ 2I = π
3
π
6
ln(tan x .cot x ) dx = 0 =⇒ I = 0
Tìmnguyênhàm I =
dx
sin3 x +cos3 x
Bài 21
Li gii:
Tacó 1
sin3x +
cos3x = (sin x +cos x )
(sin x +
cos x )2(1−
sin x cos x )= (sin x +cos x )
(1
+sin2x )(1
−sin x cos x )
Đt t = sin x −cos x , sin x cos x = 1− t 2
2,dt = (cos x + sin x ) dx
I =
dt
(2− t 2)
1− 1− t 2
2
= 2
dt
(2− t 2)(1+ t 2)= 2
3
1
2− t 2 + 1
1+ t 2
dt
I = 2
3
dt
2− t 2 + 2
3
dt
1+ t 2
Tínhtíchphân I =0
−π
4
sin4x
(1+ sin x )(1+cos x )dx
Bài 22
Li gii:
2(1+ sin x )(1+cos x )= (sin x +cos x +1)2 = 4sin2x (cos x + sin x )(cos x − sin x )
(sin x +cos x +1)2
Đt t =cos x + sin x , sin 2x = t 2 −1, dt = (cos x − sin x ) dx , x = −π
4, t =0, x = 0, t =1
5
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I =1
0
4(t 2−1)t
(t +1)2 dt =4
1
0
t 2 − t
t +1dt =4
1
0
t −2+ 2
t +1
dt
I =
2t 2 −8t +8ln(t +1) 1
0= 2(4ln2−3)
Tínhtíchphân I =
3
1 3
dx
1+x 2+ x 98+ x 100
Bài 23
Li gii:
I =
3
1 3
dx
(1+x 2)(1+x 98)=x = 1
x
3
1 3
dx
x 2
1+ 1
x 2
1+ 1
x 98
=
3
1 3
x 98 dx
(x 2 +1)(x 98+1)
=⇒ I = 1
2
3
1 3
dx
1+ x 2
Tìmnguyênhàm I =
x 2−3x + 5
47 (2x
+1)4
dx
Bài 24
Li gii:
I = 1
4
4x 2−12x +5
(2x +1)4
7
dx
I = 1
8
(2x +1)
2 −8(2x +1)+12
(2x +1)−4
7 d(2x +1)
I = 1
8
(2x +1)
107 −8(2x +1)
37 +12(2x +1)
−47
d(2x +1)
I = 7
136(2x +1)
177 − 7
10(2x +1)
107 + 9
14(2x +1)
37 +C
Tìmnguyênhàm I =
2x 3+5x 2 −11x +4
(x +1)30 dx
Bài 25
Li gii:
I =
2(x +1)3 − (x +1)2 −15(x +1)+18
(x +1)30 dx
=
2(x +1)−27− (x +1)
−28−15(x +1)−29+18(x +1)
−30 dx
=− 1
13(x +1)26 + 1
27(x +1)27 + 15
28(x +1)28 − 18
29(x +1)29 +C
Tìmnguyênhàm I =
x 3−3x 2 +4x −9
(x −2)15 dx
Bài 26
Li gii:
I =
(x −2)3 +3(x −2)2 +4(x −2)+3
(x −2)15 dx
=
(x −2)−12+3(x −2)
−13+4(x −2)−14+3(x −2)
−15 dx
=− 1
11(x −2)11 − 1
4(x −2)12 − 4
13(x −2)13 − 3
14(x +1)14 +C
Tìmnguyênhàm I =(x −1)2(5x +2)15 dx
Bài 27
Li gii: Ta có
6
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25(x −1)2 = 25x 2−50x +25= 25x 2+20x +4−70x −28+49= (5x +2)2−14(5x +2)+49
NênI = 1
25
(5x +2)
17−14(5x +2)16+49(5x +2)
15 dx
I = 1
25
(5x +2)18
90− 14(5x +2)17
85+ 49(5x +2)16
80
+C
Tínhtíchphân I =8
4
x 2 −16
x dx
Bài 28
Li gii:
Đt x = 4
sin t , dx = −4cos t
sin2t dt ,
4
sin t
2
−16= 4cot t x = 4, t = π
2; x = 8, t = π
6
Ta đưc
I = π
6
π
2
4cot t
4
sin t
−4cos t
sin2t dt =4
π
2
π
6
cot2
t dt =4
π
2
π
6
(1+cot2
t −1) dt
=4(
−cot t
−t )
π
2
π
6 =4
3
+
4π
3
Tínhtíchphân I =1
1 3
(1+x 2)5
x 8 dx
Bài 29
Li gii:
Đt x = tan t , dx = dt
cos2t ,
(1+ x 2)
5 =
1
cos10t , x = 1
3, t = π
6, x = 1, t = π
4
Ta đưc
I =
π
4
π
6
1
cos10t
tan8t
dt
cos2t =π
4
π
6
d(sin t )
si n 8t
dt =
1
7sin
7t
π
4
π
6 =128−8
2
7
Tínhtíchphân I =2
1
x −
x 2−2x +2
x +
x 2−2x +2
dx
x 2−2x +2
Bài 30
Li gii: Đt x =u +1, dx = du , x = 1,u = 0, x = 2,u = 1
Ta đưc
I =1
0
u +1−
u 2+1
x +1
x 2 +1
du
u 2+1=1
0
du
u 2 +1−1
0
2 du u 2+1(u +
u 2+1+1)
=1
0du
u 2 +1−1
+
2
12 dt
t (t +1)( vi t =u +
u 2 +1, dt =
u
2
+1+u u 2 +1
du )
= arctanu
1
0−2 ln
t
t +1
1+
2
1= π
4− ln 2
7
