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部分分式积分法 .
部分分式积分法 ，即通过将原函数 拆分为部分分式 来简化积分 步骤，是计算积分时的一个常用技巧。任何有理 函数都可拆分为多个多项式 和部分分式的和，每个部分分式中的分子 次数小于分母 ，然后根据积分表 及利用其他积分技巧，将每个部分分式积分，就得到原函数的积分。
以下是一个简单的例子。计算
∫
10
x
2
+
12
x
+
20
x
3
−
8
d
x
{\displaystyle \int {10x^{2}+12x+20 \over x^{3}-8}\,dx}
10
x
2
+
12
x
+
20
x
3
−
8
=
10
x
2
+
12
x
+
20
(
x
−
2
)
(
x
2
+
2
x
+
4
)
=
A
x
−
2
+
B
x
+
C
x
2
+
2
x
+
4
{\displaystyle {10x^{2}+12x+20 \over x^{3}-8}={10x^{2}+12x+20 \over (x-2)(x^{2}+2x+4)}={A \over x-2}+{Bx+C \over x^{2}+2x+4))
通分得到：
10
x
2
+
12
x
+
20
=
A
(
x
2
+
2
x
+
4
)
+
(
B
x
+
C
)
(
x
−
2
)
{\displaystyle 10x^{2}+12x+20=A(x^{2}+2x+4)+(Bx+C)(x-2)\,}
整理，原式变为：
10
x
2
+
12
x
+
20
=
(
A
+
B
)
x
2
+
(
2
A
−
2
B
+
C
)
x
+
(
4
A
−
2
C
)
{\displaystyle 10x^{2}+12x+20=(A+B)x^{2}+(2A-2B+C)x+(4A-2C)\,}
因此，
A
+
B
=
10
{\displaystyle A+B=10\,}
2
A
−
2
B
+
C
=
12
{\displaystyle 2A-2B+C=12\,}
4
A
−
2
C
=
20
{\displaystyle 4A-2C=20\,}
解方程组，得到：
A
=
7
{\displaystyle A=7\,}
B
=
3
{\displaystyle B=3\,}
C
=
4
{\displaystyle C=4\,}
所以：
10
x
2
+
12
x
+
20
x
3
−
8
=
7
x
−
2
+
3
x
+
4
x
2
+
2
x
+
4
{\displaystyle {10x^{2}+12x+20 \over x^{3}-8}={7 \over x-2}+{3x+4 \over x^{2}+2x+4))
即：
∫
10
x
2
+
12
x
+
20
x
3
−
8
d
x
=
∫
(
7
x
−
2
+
3
x
+
4
x
2
+
2
x
+
4
)
d
x
=
∫
7
x
−
2
d
x
+
∫
3
x
+
4
x
2
+
2
x
+
4
d
x
{\displaystyle \int {10x^{2}+12x+20 \over x^{3}-8}\,dx=\int ({7 \over x-2}+{3x+4 \over x^{2}+2x+4})\,dx=\int {7 \over x-2}\,dx+\int {3x+4 \over x^{2}+2x+4}\,dx}
利用换元积分法 ，将
x
−
2
{\displaystyle x-2\,}
x
2
+
2
x
+
4
{\displaystyle x^{2}+2x+4\,}
∫
10
x
2
+
12
x
+
20
x
3
−
8
d
x
{\displaystyle \int {10x^{2}+12x+20 \over x^{3}-8}\,dx}
=
7
ln
|
x
−
2
|
+
∫
3
2
(
2
x
+
2
)
+
1
x
2
+
2
x
+
4
d
x
{\displaystyle =7\ln |x-2|+\int (({\frac {3}{2))(2x+2)+1} \over x^{2}+2x+4}\,dx}
=
7
ln
|
x
−
2
|
+
3
2
∫
2
x
+
2
x
2
+
2
x
+
4
d
x
+
∫
1
(
x
+
1
)
2
+
3
d
x
{\displaystyle =7\ln |x-2|+{\frac {3}{2))\int {2x+2 \over x^{2}+2x+4}\,dx+\int {1 \over (x+1)^{2}+3}\,dx}
=
7
ln
|
x
−
2
|
+
3
2
ln
|
x
2
+
2
x
+
4
|
+
1
3
arctan
(
x
+
1
3
)
+
C
{\displaystyle =7\ln |x-2|+{\frac {3}{2))\ln |x^{2}+2x+4|+{\frac {1}{\sqrt {3))}\arctan({x+1 \over {\sqrt {3))})+C}
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