题目 08
$$
I=\int \frac{x}{x^{3}-x^{2}+x-1} \mathrm{~ d} x = ?
$$
解析 08
用十字相乘法进行拆分：
$$
x^{3}-x^{2}+x-1 \Rightarrow
$$
$$
(x \quad \quad) \cdot\left(x^{2} \quad \quad \right) \Rightarrow x^{3} \Rightarrow
$$
$$
(x-1) \cdot\left(x^{2}\quad \quad \right) \Rightarrow x^{3}-x^{2} \Rightarrow
$$
$$
(x-1) \cdot\left(x^{2}+1\right) \Rightarrow x^{3}-x^{2}+x-1
$$
于是：
$$
\frac{x}{x^{3}-x^{2}+x-1}=\frac{x}{(x-1)\left(x^{2}+1\right)}=
$$
待定系数变乘法为加减法：
$$
\frac{A}{x-1}+\frac{B x+ C}{x^{2}+1} \Rightarrow
$$
$$
\frac{A\left(x^{2}+1\right)+(B x+ C)(x-1)}{(x-1)\left(x^{2}+1\right)}=\frac{x}{(x-1)\left(x^{2}+1\right)} \Rightarrow
$$
$$
A x^{2}+A+B x^{2}-B x+ C x-C=x \Rightarrow
$$
$$
\left\{\begin{array}{l}A+B=0 \\ C-B=1 \\ A-C=0\end{array}\right. \Rightarrow \left\{\begin{array}{l}A+B=0 \\ A-B=1\end{array} \Rightarrow\right. \left\{\begin{array}{l}A=\frac{1}{2} \\ B=\frac{-1}{2} \\ C=\frac{1}{2}\end{array}\right.
$$
于是：
$$
I=\int\left(\frac{\frac{1}{2}}{x-1}+\frac{\frac{-1}{2} x+\frac{1}{2}}{x^{2}+1}\right) \mathrm{~ d} x
$$
$$
I=\frac{1}{2} \int \frac{1}{x-1} \mathrm{~ d} x+\frac{1}{2} \int \frac{1-x}{x^{2}+1} \mathrm{~ d} x \Rightarrow
$$
$$
I=\frac{1}{2} \ln |x-1|+\frac{1}{2}\left[\int \frac{1}{x^{2}+1} \mathrm{~ d} x-\int \frac{x}{x^{2}+1} \mathrm{~ d} x\right]=
$$
$$
I=\frac{1}{2} \ln |x-1|+\frac{1}{2} \arctan x-\frac{1}{2} \int \frac{x}{x^{2}+1} \mathrm{~ d} x .
$$
$$
I=\frac{1}{2} \ln |x-1|+\frac{1}{2} \arctan x-\frac{1}{4} \ln \left(x^{2}+1\right)+ C
$$