天津网站推广有哪些论坛seo教程
前置知识：有理函数的不定积分
习题
计算 ∫ x 3 + 1 x 4 − 3 x 3 + 3 x 2 − x d x \int \dfrac{x^3+1}{x^4-3x^3+3x^2-x}dx ∫x4−3x3+3x2−xx3+1dx
解：
\qquad 将被积函数的分母因式分解得
x 4 − 3 x 3 + 3 x 2 − x = x ( x − 1 ) 3 x^4-3x^3+3x^2-x=x(x-1)^3 x4−3x3+3x2−x=x(x−1)3
设被积函数有分解式
x 3 + 1 x 4 − 3 x 3 + 3 x 2 − x = A x + B x − 1 + C ( x − 1 ) 2 + D ( x − 1 ) 3 \dfrac{x^3+1}{x^4-3x^3+3x^2-x}=\dfrac Ax+\dfrac{B}{x-1}+\dfrac{C}{(x-1)^2}+\dfrac{D}{(x-1)^3} x4−3x3+3x2−xx3+1=xA+x−1B+(x−1)2C+(x−1)3D
将上式右端通分合并，分母相等，分子也应相等，得
x 3 + 1 = ( A + B ) x 3 + ( − 3 A − 2 B + C ) x 2 + ( 3 A + B − C + D ) x − A x^3+1=(A+B)x^3+(-3A-2B+C)x^2+(3A+B-C+D)x-A x3+1=(A+B)x3+(−3A−2B+C)x2+(3A+B−C+D)x−A
可列方程组
{ A + B = 1 − 3 A − 2 B + C = 0 3 A + B − C + D = 0 − A = 1 \begin{cases} A+B=1 \\ -3A-2B+C=0 \\ 3A+B-C+D=0 \\ -A=1 \end{cases} ⎩ ⎨ ⎧A+B=1−3A−2B+C=03A+B−C+D=0−A=1
解得
{ A = − 1 B = 2 C = 1 D = 2 \begin{cases} A=-1 \\ B=2 \\ C=1 \\ D=2 \end{cases} ⎩ ⎨ ⎧A=−1B=2C=1D=2
所以
\qquad 原式 = − ∫ 1 x d x + 2 ∫ 1 x − 1 d x + ∫ 1 ( x − 1 ) 2 d x + 2 ∫ 1 ( x − 1 ) 3 d x =-\int \dfrac 1xdx+2\int \dfrac{1}{x-1}dx+\int \dfrac{1}{(x-1)^2}dx+2\int \dfrac{1}{(x-1)^3}dx =−∫x1dx+2∫x−11dx+∫(x−1)21dx+2∫(x−1)31dx
= − ln ∣ x ∣ + 2 ln ∣ x − 1 ∣ − 1 x − 1 − 1 ( x − 1 ) 2 + C \qquad\qquad =-\ln|x|+2\ln|x-1|-\dfrac{1}{x-1}-\dfrac{1}{(x-1)^2}+C =−ln∣x∣+2ln∣x−1∣−x−11−(x−1)21+C