凑微分法与分部积分法

= −1 ⇒
B
=1 .
A − C = 2
C = −1
∫ ∫ ∫ 故
(x
2x2 − x + 2 −1)(x2 + x +1)
=
1+ x −1
x
2
x −1 +x+
1
.
⇒
(
x
2x2 −1)(
−x+2 x2 + x +
1)
dx
=
1 dx + x −1
x
2
x −1 +x+
dx 1
=
ln
x
−1
+
1 2
∫
2x − x2 + x
+
∫
x
1 −
dx 3
= − ln x +1 + ln x − 3 + C = ln x − 3 + C. x +1
2018/12/13
Edited by Lin Guojian
1
2x
例
:
∫
x2
dx. −1
解:设
2x x2 −1
=
2x (x −1)(x
+ 1)
=
A+ x −1
B. x +1
由于 A + B = A(x +1) + B(x −1) ⇒ 2x = A(x +1) + B(x −1)
2
22
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Edited by Lin Guojian
10
例: ∫ ln xdx.
解
:
∫
ln
xdx
=
x
ln
x
−
∫
xd
ln
x
=
x
ln
x
−
∫
x
⋅
1 x
dx
=x
ln
x
−
∫
dx
=x
ln
x
−
x
+
C.
例: ∫ arcsin xdx.
解 : ∫ arcsin xdx = x arcsin x − ∫ xd arcsin x = x arcsin x − ∫ x ⋅
2018/12/13
Edited by Lin Guojian
8
∫ 例: x3exdx.
∫ ∫ ∫ 解 : x3exdx = x3dex = x3ex − 3 x2exdx
∫ = x3ex − 3 x2dex ∫ = x3ex − 3x2ex + 6 xexdx ∫ = x3ex − 3x2ex + 6 xdex ∫ = x3ex − 3x2ex + 6xex − 6 exdx
⇒ 4 = A(x − 3) + B(x +1) = ( A + B)x − 3A + B
⇒
−
A+B =0 3A+ B = 4
⇒
A = −1 B =1
⇒
(x
4 + 1)( x
−
3)
=
−1 x +1
+
x
1 −
. 3
故∫
(
x
+
4 1)(x
−
3)
dx
=
∫
(
−1 x +1
+
x
1 −
)dx 3
=
−∫
x
1 dx +1
即∫ u(x)dv(x) = u(x)v(x) − ∫ v(x)du(x), (∫ udv = uv − ∫ vdu).
2018/12/13
Edited by Lin Guojian
6
注 : 分部积分公式中,左边积分∫ udv通常不易求出,右边积分∫ vdu容易求出.
注 : 分部积分公式是函数乘积的导数公式或微分公式的逆运算公式.
∫ = x2 sin(x2 +1) − sin(x2 +1)dx2 ∫ = x2 sin(x2 +1) − sin(x2 +1)d (x2 +1)
= x2 sin(x2 +1) + cos(x2 +1) + C.
2018/12/13
Edited by Lin Guojian
13
例 : ∫ x2 ln(x +1)dx.
使用分部积分法的常见题型 :
∫ ∫ ∫ 1: 形如 xµexdx, xµ cos xdx, xµ sin xdx.
选择u(x) = xµ , v′(x) = ex , cos x,sin x.
2 : 形如∫ xµ ln xdx, ∫ xµ arcsin xdx, ∫ xµ arctan xdx, ∫ xµ arccos xdx, ∫ xµarc cot xdx.
例
:
∫
(
x
+
4 1)(
x
−
dx. 3)
解:设
4
= A+ B.
(x +1)(x − 3) x +1 x − 3
由于 A + B = A(x − 3) + B(x +1) ⇒
4
= A(x − 3) + B(x +1)
x +1 x − 3 (x +1)(x − 3)
(x +1)(x − 3) (x +1)(x − 3)
选择u(x) = ln x, arcsin x, arctan x, arccos x, arc cot x, v′(x) = xµ .
2018/12/13
Edited by Lin Guojian
7
例: ∫ x sin xdx.
解 : ∫ x sin xdx = −∫ xd cos x = −x cos x + ∫ cos xdx = −x cos x + sin x + C.
4
4
4
16
例: ∫ x arctan xdx.
∫ ∫ ∫ 解 : x arctan xdx = 1 arctan xdx2 = 1 x2 arctan x − 1 x2 ⋅ 1 dx
2
2
2
x2 +1
∫ = 1 x2 arctan x − 1 (1− 1 )dx
2
2
x2 +1
= 1 x2 arctan x − 1 x + 1 arctan x + C.
例: ∫ x2 cos xdx.
解 : ∫ x2 cos xdx = ∫ x2d sin x = x2 sin x − ∫ 2x sin xdx = x2 sin x + 2∫ xd cos x
= x2 sin x + 2x cos x − 2∫ cos xdx
= x2 sin x + 2x cos x − 2sin x + C.
1 dx
1− x2
∫ = x arcsin x + 1 1 d (1− x2 ) 2 1− x2
= x arcsin x + 1− x2 + C.
2018/12/13
Edited by Lin Guojian
11
例: ∫ x(ln x)2 dx.
∫ ∫ ∫ 解 : x(ln x)2 dx = 1 (ln x)2 dx2 = 1 (ln x)2 ⋅ x2 − 1 x2d (ln x)2
2
2
2
∫ ∫ = 1 x2 (ln x)2 − 1 x2 ⋅ 2 ln x ⋅ 1 dx = 1 x2 (ln x)2 − x ln xdx
2
2
x2
∫ ∫ = 1 x2 (ln x)2 − 1 ln xdx2 = 1 x2 (ln x)2 − 1 x2 ln x + 1 x2 ⋅ 1 dx
2
2
2
2
2x
解
:
∫
x2
ln(x
+
1)dx
=
1 3
∫
ln(x
+
1)dx3
∫ = 1 x3 ln(x +1) − 1 x3 dx
3
3 x +1
∫ = 1 x3 ln(x +1) − 1 [x2 − x +1− 1 ]dx
3
3
x +1
= 1 x3 ln(x +1) − 1 [1 x3 − 1 x2 + x − ln(x +1)] + C.
x −1 x +1 (x −1)(x +1)
(x −1)(x +1) (x −1)(x +1)
⇒ 2x = A(x +1) + B(x −1) = ( A + B)x + A − B
⇒
A+ B
A
−
B
= =
2 0 (x −1)(x +1)
=
1+ x −1
1. x +1
故
∫
(
x
5
二、分部积分法
定理5.3(分部积分法) : 设u = u(x), v = v(x)有连续的导数(即导函数连续).
则有分部积分公式 :
∫ u(x)v′(x)dx = u(x)v(x) − ∫ u′(x)v(x)dx.
或∫ u(x)dv(x) = u(x)v(x) − ∫ v(x)du(x), (∫ udv = uv − ∫ vdu).
3
33 2
= − 1 x3 + 1 x2 − 1 x + 1 (x3 +1) ln(x +1) + C. 9 6 33
2018/12/13
Edited by Lin Guojian
14
∫ 例 : 4x3 ln(x2 +1)dx.
∫ ∫ 解 : 4x3 ln(x2 +1)dx = ln(x2 +1)dx4 ∫ = x4 ln(x2 +1) − 2 x5 dx
2x −1)(x
+
1)
dx
=
∫
(
x
1 −1
+
x
1 +
)dx 1
=
∫
x
1 dx −1
+
∫
x
1 +
dx 1
= ln x −1 + ln x +1 + C = ln x2 −1 + C.
2018/12/13
Edited by Lin Guojian
2
∫ 例 :
2(
x
x2 +8 −1)(x +
2)
2
dx.
+
C.
2018/12/13
Edited by Lin Guojian
3
2x2 − x + 2
例
:
∫
(
x
−1)(
x2
+
x
+
dx. 1)
2x2 − x + 2 解 : 设 (x −1)(x2 + x +1)
=
A+ x −1
Bx + C
x2
+
x
+
. 1
由于 A + x −1
Bx + x2 + x
C +
1
= x3ex − 3x2ex + 6xex − 6ex + C.
2018/12/13
Edited by Lin Guojian
9
例: ∫ x3 ln xdx.
∫ ∫ ∫ 解 : x3 ln xdx = 1 ln xdx4 = 1 x4 ln x − 1 x4 1 dx
4
4
4x
∫ = 1 x4 ln x − 1 x3dx = 1 x4 ln x − 1 x4 + C.
x8
1 (x8
d ( x8 ) + 1)
=
1 8
∫
1 du u(u +1)
=
1 8
∫[
1 u
−
u
1 ]du +1
2018/12/13
=
1 8
∫
1 u
du
−
1 8
∫
u
1d +1
(u
+ 1)
=
1 8
ln
u
−
1 8
ln
u
+1
+
C
= 1 ln x8 − 1 ln x8 +1 + C.
8
8
Edited by Lin Guojian
3 arctan 2
3(x + 1) 2 + C.
22
23
3
2018/12/13
Edited by Lin Guojian
4
例
:
∫
x(
1 x8
+
dx. 1)
∫ ∫ ∫ 解 :
1
x(x8
dx + 1)
=
x8
(
x7 x8
+
dx 1)
=
1 8
x8
(
1 x8
d + 1)
(x8
).
令u = x8.
则
1 8
∫
2 dx +1
=
ln
x
−1
+
1 2
∫
2x x2
+1− 3dx + x +1
=
ln
x
−1
+
1 2
∫
2x x2 +
+1 dx x +1
−
3 2
∫
x2
1 +x
dx +1
∫ ∫ = ln x −1 + 1
2
x2
1 +x
+
d 1
(
x2
+
x
+
1)
−
3 2
1 (x + 1)2 + (
d(x + 1)
3 )2
2
= ln x −1 + ln(x2 + x +1) − 3 2
=
A( x 2
+ x +1) + (Bx + C)(x −1)
(x −1)(x2 + x +1)
.
⇒ 2x2 − x + 2 = A(x2 + x +1) + (Bx + C)(x −1) = ( A + B)x2 + ( A − B + C)x + A − C.
A+B = 2
A=1
⇒
A
−
B
+
C
证 :由导数的乘法公式 : [u(x)v(x)]′ = u′(x)v(x) + u(x)v′(x) ⇒ u(x)v′(x) = [u(x)v(x)]′ − u′(x)v(x).