30 BÀI TOÁN TÍCH PHÂN
Z 6x3+ 8x + 1 (3x2+ 4)px2+ 1 dx
Lời giải:
Ta có 6x
3
+ 8x + 1
3x2+ 4 = 2x +
1
3x2+ 4
=⇒ I =
Z µ
3x2+ 4
p
x2+ 1dx =
p
x2+ 1dx +
(3x2+ 4)px2+ 1dx
TínhI1=
p
x2+ 1 dx
Đặtpx2+ 1 = t , x2+ 1 = t2, 2tdt = 2xdx =⇒ I1= 2
Z tdt
t = 2t = 2px2+ 1
TínhI2=
(3x2+ 4)px2+ 1.dx
Đặtt =
p
x2+ 1
x , xt =px2+ 1, x2t2= x2+ 1, x2= 1
t2− 1, 3x
2
+ 4 =4t
2
− 1
t2− 1
xdx = − t dt
(t2− 1)2, dx
xt = − t dt
(t2− 1)2x2t,
dx
p
x2+ 1=
dt
1 − t2
I2=
Z t2− 1
4t2− 1
dt
1 − t2=
1 − 4t2 =1
2
Z µ 1
2t + 1−
1
2t − 1
¶
dt =1
4ln
2t + 1 2t − 1=
1
4ln
2p
x2+ 1 + x
2p
x2+ 1 − x
4ln
2p
x2+ 1 + x
2p
x2+ 1 − x +C
sin x +p3 cos x dx
Lời giải:
Dùng pp hệ số bất địnhcos2x = (a sin x + b cos x)(sin x +p3 cos x) + c(sin2x + cos2x)
cos2x =
Ã
−1
4 sin x +
p 3
4 cos x
!
(sin x +p3 cos x) +1
4=−1
4 (sin x −p3 cos x)(sin x +p3 cos x) +1
4
I =
Z −1
4 (sin x −p3 cos x)(sin x +p3 cos x) +14
=−1 4
Z
(sin x −p3 cos x)dx +1
4
sin x +p3 cos x dx
=1
4(cos x +p3 sin x) +1
4
sin x +p3 cos x dx
Ta tínhJ =1
4
sin x +p3 cos x=1
8
cos(x − π6)=1
8
Z cos(x − π6)
1 − sin2(x − π6)dx
Đặtt = sin(x − π6) =⇒ dt = cos(x − π6)dx
=⇒ J =1
8
Z dt
1 − t2= 1
16
Z µ 1
t + 1−
1
t − 1
¶
dt = 1
16ln
t + 1
t − 1=
1
16ln
sin(x − π6) + 1
sin(x − π6) − 1
4(cos x +p3 sin x) + 1
16ln
sin(x − π6) + 1
sin(x − π6) − 1+C
Z x3+ x2
4
p
4x + 5 dx
Lời giải:
Trang 2I =
Z x3+ x2
4
p
4x + 5 dx =
Z x4+ x3
4
p
4x5+ 5x4 dx
= 1 20
Z
¡4x5
+ 5x4¢−
1
d(4x5+ 5x4) = 1
15
4
p
(4x5+ 5x4)3+C
Z ³
cos 2x +p2 cos³x + π
4
´´
e sin x+cos x+1dx
Lời giải:
Ta có cos 2x +p2 cos¡x + π4¢ = (cos x − sin x)(sin x + cos x + 1)
I =
Z
(cos x − sin x)(sin x + cos x + 1)e sin x+cos x+1dx
=
Z
(sin x + cos x + 1)e sin x+cos x+1d(sin x + cos x + 1)
=
Z
(sin x + cos x + 1)d¡e sin x+cos x+1¢
=(sin x + cos x + 1)e sin x+cos x+1−
Z
e sin x+cos x+1d(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
Z
3
p
3x − x3dx
Lời giải:
Đặtt =
3
p
3x − x3
x =⇒ x2= 3
t3+ 1 =⇒ 2xdx = −9t
2dt
(t3+ 1)2
I =1
2
Z p3
3x − x3
x 2xdx =−9
2
Z t3dt
(t3+ 1)2=
3 2
Z
td
t3+ 1
¶
2(t3+ 1)−
3 2
Z dt
t3+ 1
TínhJ =
Z dt
t3+ 1=
(t + 1)[(t + 1)2− 3(t + 1) + 3]=
1
2(ln 3(1 − t) − 2ln3t + ln(1 + t))
VậyI =1
2x
3
p
3x − x3−3
4
Ã
ln 3
Ã
1 −
3
p
3x − x3
x
!
− 2 ln 3
3
p
3x − x3
x + ln
Ã
1 +
3
p
3x − x3
x
!!
+C
x4+ 4x3+ 6x2+ 7x + 4dx
Lời giải:
Tổng các hệ số bậc chẵn bằng tổng các hệ số bậc lẻ nên đa thức ở mẫu nhậnx = −1làm nghiệm
I =
(x + 1)[(x + 1)3+ 3]=
1 3
Z (x + 1)3+ 3 − (x + 1)3 (x + 1)[(x + 1)3+ 3] dx =1
3
·Z dx
x + 1−
Z (x + 1)2 (x + 1)3+ 3dx
¸
=1
3
·
ln |x + 1| −1
3
Z d((x + 1)3)
(x + 1)3+ 3
¸
=1
3ln |x + 1| −1
9ln |(x + 1)3+ 3| +C
Z 1 0
x ln³x +p1 + x2´
x +p1 + x2 dx
Lời giải:
Đặtu = ln(x +px2+ 1), dv = xdx
x +px2+ 1= x(
p
x2+ 1 − x)dx
Suy ra du =
1 +p x
x2+ 1
x +px2+ 1 dx =pdx
x2+ 1, v =1
2
Z
(1 + x2)1d(1 + x2) −
Z
x2dx =1
3[(1 + x2)3− x3]
I =1
3[(1 + x2)3− x3] ln(x +p1 + x2)¯¯
1
0−1 3
Z 1
0 [(1 + x2)3− x3]pd x
1 + x2
Trang 3Mà J =
Z
[(1 + x2)3− x3]pd x
1 + x2=
Z d x
1 + x2−
Z x3d x
p
1 + x2 = arctan x −1
3(x
2
− 2)px2+ 1
3[(1 + x2)3− x3] ln(x +p1 + x2)¯¯
1
0−1
3arctan x
¯
¯
1
0+1
9(x
2− 2)px2+ 1¯¯
1 0
3(
p
8 − 1)ln(1 +p2) − π
12+1
9(2 +p2)
Z 1 2
0
x ln 1 + x
1 − x dx
Lời giải:
Vớiu = ln 1 + x
1 − x, dv = xdx nên du = 2
1 − x2 dx, v =1
2x
2
I =1
2x
2ln1 + x
1 − x
¯
¯
1
0−
Z 1
0
x2
1 − x2 dx =1
8ln 3 +
Z 1
0
1 − x2− 1
1 − x2 dx
=1
8ln 3 +1
2−1 2
Z 1
0
µ 1
1 + x+
1
1 − x
¶
dx =1
8ln 3 +1
2−1
2ln
1 + x
1 − x
¯
¯
1
0=1
2−3
8ln 3
Z π
0
e −x cos 2xdx
Lời giải:
I =
Z π
0
e −x cos 2xdx = −
Z π
0 cos 2xd(e −x ) = −e −x cos 2x
¯
¯
π
0− 2
Z π
0
e −x sin 2xdx
= e −π+ 1 + 2
Z π
0
sin 2xd(e −x ) = e −π + 1 + 2e −x sin 2x¯¯π
0− 4
Z π
0
e −x cos 2xdx = 1
5(e
−π+ 1)
Z
p 3 0
x5+ 2x3
p
x2+ 1 dx
Lời giải:
I =
Z p 3 0
x(x4+ 2x2) p
x2+ 1 dx =
Z p 3 0
(x4+ 2x2)d(px2+ 1)
I = (x4+ 2x2)px2+ 1
¯
¯
p 3
0 − Z
p 3 0
p
x2+ 1d(x4+ 2x2)
Z p
x2+ 1d(x4+ 2x2) =
Z
4x(x2+ 1)px2+ 1dx = 4
Z x(x2+ 1)2 p
x2+ 1 dx
= 4
Z (px2+ 1)4d(px2+ 1) =4
5(x
2
+ 1)2px2+ 1
Nên I = (x4+ 2x2)px2+ 1
¯
¯
p 3
0 −4
5(x
2
+ 1)2px2+ 1
¯
¯
p 3 0
Z e
1
1 + x2ln x
x + x2ln x dx
Lời giải:
I =
Z e
1
1 + x2ln x
x + x2ln x dx =
Z e
1
1
x2+ ln x
1
x + ln x
dx =
Z e
1
1
x + ln x
1
x + ln x
dx +
Z e
1
1
x2−1
x
1
x + ln x
dx
=
Z e
1
dx −
Z e
1
dµ 1
x + ln x
¶
1
x + ln x
= x¯¯ e
1− lnµ 1
x + ln x
¶¯
¯
e
1= e − 1 − lnµ 1
e+ 1
¶
Trang 4Tìm nguyên hàm I =
Z
2(1 + ln x) + x ln x(1 + ln x)
Lời giải:
Đặt u = 1 + x ln x =⇒ du = (1 + ln x) dx
I =
Z (2 + x ln x)(1 + ln x)
1 + x ln x dx =
Z u + 1
u du = u + ln|u| +C = 1 + x ln x + ln|1 + x ln x| +C
Z π
4
0
x2(x2sin 2x + 1) − (x − 1)sin2x cos x(x2sin x + cos x) dx
Lời giải:
I =
Z x4sin 2x + x2− (x − 1) sin 2x
x2sin x cos x + cos2x dx =
Z π
4
0
2x4sin 2x + 2x2− 2x sin x + 2 sin 2x
x2sin 2x + cos2x + 1 dx
=
Z π
4
0
2x2(x2sin 2x + cos2x + 1) − (x2sin 2x + cos2x + 1)0
=
Z π
4
0
2x2dx −
Z π
4
0
d(x2sin 2x + cos2x + 1)
x2sin 2x + cos2x + 1
=2
3x
3¯
¯
π
4
0 − ln |x2sin 2x + cos2x + 1|¯¯
π
4
0 = π3
96+ ln 2 − ln
µπ2
16+ 1
¶
Z (x2+ 1) + (x3+ x ln x + 2) ln x
Lời giải:
I =
Z (x2+ ln x) + x ln x(x2+ ln x) + (1 + ln x)
Z (x2+ ln x)(1 + x ln x) + (1 + ln x)
=
Z
(x2+ ln x)dx +
Z d(1 + x ln x)
1 + x ln x =
1
3.x
3
+ x ln x − x + ln |1 + x ln x| +C
Z x2(x2sin2x + sin2x + cos x) + sin x(2x − 1 − sin x) + 1
Lời giải:
Vìx2(x2sin2x + sin2x + cos x) + sin x(2x − 1 − sin x) + 1 = (x2sin x + cos x)2+ (x2sin x + cos x)0
I =
Z
(x2sin x + cos x)dx +
Z d(x2sin x + cos x)
x2sin x + cos x =
Z
x2sin xdx + sin x + ln|x2sin x + cos x|
TínhJ =
Z
x2sin xdx = −
Z
x2d(cos x) = −x2cos x + 2
Z
x cos xdx = −x2cos x + 2
Z
xd(sin x)
J = −x2cos x + 2x sin x − 2
Z
sin xdx = −x2cos x + 2x sin x + 2cos x
Vậy I = −x2cos x + 2x sin x + 2cos x + sin x + ln|x2sin x + cos x| +C
Tìm nguyên hàm I =
Z ³
x(x + 2)(3sin x − 4sin3x) + 2cos x(cos x − 2sin x) + 3x2cos 3x − 1´e xdx
Lời giải:
³
x(x + 2)(3sin x − 4sin3x) + 2cos x(cos x − 2sin x) + 3x2cos 3x − 1´e x
=³x2sin 3x + (x2sin 3x)0+ cos 2x + (cos 2x)0´e x
=⇒ I = (x2sin 3x + cos2x)e x
Z 2x4ln2x + x ln x(x3+ 1) + x − x12
Lời giải:
Trang 52x6ln2x + x6ln x + x3ln x + x3− 1
x2+ x5ln x =2[(x
3ln x)2− 1] + x3(x3ln x + 1) + (x3ln x + 1)
x2(1 + x3ln x)
=(x
3ln x + 1)(2x3ln x + x3− 1)
x2(1 + x3ln x) = 2x ln x + x − 1
x2
Z µ
2x ln x + x − 1
x2
¶
dx =1
2x
2
+1
x+
Z
2x ln xdx =1
2x
2
+1
x+
Z
ln xd(x2)
I =1
2x
2
+1
x + x2ln x −
Z
xdx =1
x + x2ln x +C
Z
x2sin(ln x)dx
Lời giải:
Đặtx = e t, ln x = t, dx = e tdt
=⇒ I =
Z
e 3t sin tdt = −e 3t cos t +
Z
3e 3t cos tdt = −e 3t cos t + 3e 3t sin t −
Z
9e 3t sin t dt
=⇒ 10I = 3e 3t sin t − e 3t cos t =⇒ I = 1
10
³
3.e 3 ln x sin(ln x) − e 3 ln x cos(ln x)´+C
Z e x (x − 1) + 2x3+ x3(e x + x(x2+ 1))
e x x + x2(x2+ 1) dx
Lời giải:
e x (x − 1) + 2x3+ x3(e x + x(x2+ 1))
e x x + x2(x2+ 1) =
x3− 1
x +3x
2
+ e x+ 1
x3+ x + e x = x2−1
x+(x
3
+ x + e x)0
x3+ x + e x
Do đó
I = x
3
3 − ln |x| + ln |x3+ x + e x | +C
Z π
3
π
6
ln(tan x)dx
Lời giải:
I =
Z π
3
π
6
ln(tan x)dx=đổi biến(x= π2−x)Z
π
3
π
6
ln(cot x)dx =⇒ 2I =
Z π
3
π
6
ln(tan x cot x)dx = 0 =⇒ I = 0
sin3x + cos3x
Lời giải:
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)
Đặt t = sin x − cos x, sin x cos x = 1 − t
2
2 ,dt = (cos x + sin x)dx
I =
(2 − t2)
µ
1 −1 − t
2
2
¶ =2
(2 − t2)(1 + t2)=2
3
Z µ 1
2 − t2+ 1
1 + t2
¶
dt
I =2
3
Z dt
2 − t2+2
3
Z dt
1 + t2
Z 0
−π
4
sin 4x (1 + sin x)(1 + cos x) dx
Lời giải:
2(1 + sin x)(1 + cos x) = (sin x + cos x + 1)2=4 sin 2x(cos x + sin x)(cos x − sin x)
(sin x + cos x + 1)2
Đặtt = cos x + sin x, sin 2x = t2− 1, dt = (cos x − sin x)dx, x = −π
4 , t = 0, x = 0, t = 1
Trang 6I =
Z 1 0
4(t2− 1)t (t + 1)2 dt = 4
Z 1 0
t2− t
t + 1 dt = 4
Z 1
0 t − 2 + 2
t + 1 dt
I = ¡2t2
− 8t + 8 ln(t + 1)¢
¯
¯
1
0= 2(4 ln 2 − 3)
Z
p 3
1 p 3
dx
1 + x2+ x98+ x100
Lời giải:
I =
Z p 3
1 p 3
dx
(1 + x2)(1 + x98)=x=1
Z p 3
1 p 3
dx
x2
µ
1 + 1
x2
¶ µ
1 + 1
x98
¶ =
Z p 3
1 p 3
x98dx
(x2+ 1)(x98+ 1)
=⇒ I =1
2
Z p 3
1 p 3
dx
1 + x2
Z x2− 3x +5
4
7
p
(2x + 1)4 dx
Lời giải:
I =1
4
Z 4x2− 12x + 5 (2x + 1)4 d
x
I =1
8
Z
£(2x + 1)2− 8(2x + 1) + 12¤ (2x + 1)−47d(2x + 1)
I =1
8
Z h
(2x + 1)107 − 8(2x + 1)3+ 12(2x + 1)−47
i
d(2x + 1)
I = 7
136(2x + 1)177 − 7
10(2x + 1)107 + 9
14(2x + 1)37+C
Z 2x3+ 5x2− 11x + 4
(x + 1)30 dx
Lời giải:
I =
Z 2(x + 1)3− (x + 1)2− 15(x + 1) + 18
=
Z
£2(x + 1)−27− (x + 1)−28− 15(x + 1)−29+ 18(x + 1)−30¤ dx
13(x + 1)26+ 1
27(x + 1)27+ 15
28(x + 1)28− 18
29(x + 1)29+C
Z x3− 3x2+ 4x − 9 (x − 2)15 dx
Lời giải:
I =
Z (x − 2)3+ 3(x − 2)2+ 4(x − 2) + 3
=
Z
£(x − 2)−12+ 3(x − 2)−13+ 4(x − 2)−14+ 3(x − 2)−15¤ dx
11(x − 2)11− 1
4(x − 2)12− 4
13(x − 2)13− 3
14(x + 1)14+C
Z
(x − 1)2(5x + 2)15dx
Lời giải:
Ta có
Trang 725(x − 1)2= 25x2− 50x + 25 = 25x2+ 20x + 4 − 70x − 28 + 49 = (5x + 2)2− 14(5x + 2) + 49
Nên
I = 1
25
Z
(5x + 2)17− 14(5x + 2)16+ 49(5x + 2)15dx
I = 1
25
· (5x + 2)18
90 −14(5x + 2)
17
85 +49(5x + 2)
16
80
¸
+C
Z 8 4
p
x2− 16
x dx
Lời giải:
Đặtx = 4
sin t, dx = −4 cos t
sin2t dt,
s
µ 4
sin t
¶2
− 16 = 4 cot t x = 4, t = π2; x = 8, t = π6
Ta được
I =
Z π
6
π
2
4 cot t
4
sin t
−4 cos t
sin2t dt = 4
Z π
2
π
6
cot2tdt = 4
Z π
2
π
6
(1 + cot2t − 1)dt
= 4(− cot t − t )
¯
¯
π
2
π
6
= 4p3 +4π
3
Z 1
1 p 3
p
(1 + x2)5
x8 dx
Lời giải:
Đặtx = tan t,dx = dt
cos2t,
q
(1 + x2)5=
r 1 cos10t, x =p1
3, t = π
6, x = 1, t = π
4
Ta được
I =
Z π
4
π
6
r 1 cos10t
tan8t
dt
cos2t =
Z π
4
π
6
d(sin t )
si n8t dt =1
7sin
7t¯¯
π
4
π
6
=128 − 8
p 2 7
Z 2 1
x −px2− 2x + 2
x +px2− 2x + 2
dx
x2− 2x + 2
Lời giải:
Đặtx = u + 1,dx = du, x = 1,u = 0, x = 2,u = 1
Ta được
I =
Z 1 0
u + 1 −pu2+ 1
x + 1px2+ 1
du
u2+ 1=
Z 1 0
du
u2+ 1−
Z 1 0
2du
p
u2+ 1(u +pu2+ 1 + 1)
=
Z 1 0
du
u2+ 1−
Z 1+p2 1
2dt
t (t + 1) (vớit = u +
p
u2+ 1, dt =
p
u2+ 1 + u
p
u2+ 1 du)
= arctan u
¯
¯
1
0− 2 ln t
t + 1
¯
¯
1+p2
4− ln 2