题目 03
$$
I=\int \frac{3 x+6}{(x-1)^{2}\left(x^{2}+x+1\right)} \mathrm{~d} x
$$
解析 03
本题主要解题思路为：待定系数。
解法一
$$
\frac{A x+B}{(x-1)^{2}}+\frac{C x+D}{x^{2}+x+1}=\frac{3 x+6}{(x-1)^{2}\left(x^{2}+x+1\right)} \Rightarrow
$$
$$
A x^{3}+A x^{2}+A x+B x^{2}+B x+B+
$$
$$
C x^{3}+C x-2 C x^{2}+ D x^{2}+D-2 D x=3 x+6 \Rightarrow
$$
$$
\left\{\begin{array} { l }
{ A + C = 0 } \\ { A + B – 2 C + D = 0 } \\ { A + B + C – 2 D = 3 } \\ { B + D = 6 }
\end{array} \Rightarrow \left\{\begin{array}{l}
A+B+2 A+6-B=0 \\ B-2(6-B)=3
\end{array}\right.\right.
$$
$$
\Rightarrow\left\{\begin{array}{l}
A=-2 \\ B=5 \\ C=2 \\ D=1
\end{array} \Rightarrow I=\int\left(\frac{-2 x+5}{(x-1)^{2}}+\frac{2 x+1}{x^{2}+x+1}\right) \mathrm{~d} x\right. \Rightarrow
$$
$$
I=
\left(\frac{-(2 x-2)+3}{x^{2}+1-2 x}+\frac{2 x+1}{x^{2}+x+1}\right) \mathrm{~d} x \Rightarrow
$$
$$
I=\int\left[\frac{-(2 x-2)}{x^{2}+1-2 x}+\frac{3}{x^{2}+1-2 x}+\frac{2 x-1}{x^{2}+x+1}\right] \mathrm{~d} x \Rightarrow
$$
$$
I=\int\left[\frac{-(2 x-2)}{x^{2}+1-2 x}+\frac{3}{(x-1)^{2}}+\frac{2 x-1}{x^{2}+x+1}\right] \mathrm{~d} x \Rightarrow
$$
$$
I=-\ln \left|x^{2}+1-2 x\right|-3 \frac{1}{x-1}+\ln \left|x^{2}+x+1\right|+C \Rightarrow
$$
$$
I=-\ln (x-1)^{2}-\frac{3}{x-1}+\ln \left(x^{2}+x+1\right)+C \Rightarrow
$$
$$
I=-2 \ln |x-1|-\frac{3}{x-1}+\ln \left(x^{2}+x+1\right)+C
$$
解法二
$$
\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{D x+E}{x^{2}+x+1}=
$$
$$
\frac{3 x+6}{(x-1)^{2}\left(x^{2}+x+1\right)} \Rightarrow
$$
$$
\frac{A(x-1)^{2}\left(x^{2}+x+1\right)+B(x-1)\left(x^{2}+x+1\right)+(D x+E)(x-1)^{3}}{(x-1)(x-1)^{2}\left(x^{2}+x+1\right)} \Rightarrow
$$
$$
A(x-1)\left(x^{2}+x+1\right)+B\left(x^{2}+x+1\right)+(D x+E)(x-1)^{2}=3 x+6
$$
$$
A x^{3}+A x^{2}+A x-A x^{2}-A x-A+B x^{2}+B x+B +
$$
$$
D x^{3}+D x-2 D x^{2}+E x^{2}+E-2 E x=3 x+6
$$
即：
$$
\left\{\begin{array}{l}
A+D=0 \\ B-2 D+E=0 \\ B+D-2 E=3 \\ -A+B+E=6
\end{array}\right. \Rightarrow \left\{\begin{array}{l}
A=-2 \\ B=3 \\ D=2 \\ E=1
\end{array} \Rightarrow \right.
$$
$$
I=\int\left[\frac{-2}{x-1}+\frac{3}{(x-1)^{2}}+\frac{2 x+1}{x^{2}+x+1}\right] \mathrm{~d} x \Rightarrow
$$
$$
I=-2 \ln |x-1|-\frac{3}{x-1}+\ln \left(x^{2}+x+1\right)+C
$$