考研数学不定积分补充例题

题目 03

$$I=\int \frac{3x+6}{(x-1{)}^2(x^2+x+1)}\mathrm{d}x$$

解析 03

本题主要解题思路为：待定系数。

解法一

$$\frac{Ax+B}{(x-1{)}^2}+\frac{Cx+D}{x^2+x+1}=\frac{3x+6}{(x-1{)}^2(x^2+x+1)}\Rightarrow$$

$$A{x}^3+A{x}^2+Ax+B{x}^2+Bx+B+$$

$$C{x}^3+Cx-2C{x}^2+D{x}^2+D-2Dx=3x+6\Rightarrow$$

$$\{\begin{array}{l}A+C=0\\ A+B–2C+D=0\\ A+B+C–2D=3\\ B+D=6\end{array}\Rightarrow \{\begin{array}{l}A+B+2A+6-B=0\\ B-2(6-B)=3\end{array}$$

$$\Rightarrow \{\begin{array}{l}A=-2\\ B=5\\ C=2\\ D=1\end{array}\Rightarrow I=\int (\frac{-2x+5}{(x-1{)}^2}+\frac{2x+1}{x^2+x+1})\mathrm{d}x\Rightarrow$$

$$I=(\frac{-(2x-2)+3}{x^2+1-2x}+\frac{2x+1}{x^2+x+1})\mathrm{d}x\Rightarrow$$

$$I=\int [\frac{-(2x-2)}{x^2+1-2x}+\frac{3}{x^2+1-2x}+\frac{2x-1}{x^2+x+1}]\mathrm{d}x\Rightarrow$$

$$I=\int [\frac{-(2x-2)}{x^2+1-2x}+\frac{3}{(x-1{)}^2}+\frac{2x-1}{x^2+x+1}]\mathrm{d}x\Rightarrow$$

$$I=-\ln |x^2+1-2x|-3\frac{1}{x-1}+\ln |x^2+x+1|+C\Rightarrow$$

$$I=-\ln (x-1{)}^2-\frac{3}{x-1}+\ln (x^2+x+1)+C\Rightarrow$$

$$I=-2\ln |x-1|-\frac{3}{x-1}+\ln (x^2+x+1)+C$$

解法二

$$\frac{A}{(x-1)}+\frac{B}{(x-1{)}^2}+\frac{Dx+E}{x^2+x+1}=$$

$$\frac{3x+6}{(x-1{)}^2(x^2+x+1)}\Rightarrow$$

$$\frac{A(x-1{)}^2(x^2+x+1)+B(x-1)(x^2+x+1)+(Dx+E)(x-1{)}^3}{(x-1)(x-1{)}^2(x^2+x+1)}\Rightarrow$$

$$A(x-1)(x^2+x+1)+B(x^2+x+1)+(Dx+E)(x-1{)}^2=3x+6$$

$$A{x}^3+A{x}^2+Ax-A{x}^2-Ax-A+B{x}^2+Bx+B+$$

$$D{x}^3+Dx-2D{x}^2+E{x}^2+E-2Ex=3x+6$$

即：

$$\{\begin{array}{l}A+D=0\\ B-2D+E=0\\ B+D-2E=3\\ -A+B+E=6\end{array}\Rightarrow \{\begin{array}{l}A=-2\\ B=3\\ D=2\\ E=1\end{array}\Rightarrow$$

$$I=\int [\frac{-2}{x-1}+\frac{3}{(x-1{)}^2}+\frac{2x+1}{x^2+x+1}]\mathrm{d}x\Rightarrow$$

$$I=-2\ln |x-1|-\frac{3}{x-1}+\ln (x^2+x+1)+C$$