$$$\frac{1}{x \left(x^{2} + 1\right)}$$$ 的积分

求$$$\int \frac{1}{x \left(x^{2} + 1\right)}\, dx$$$。

设$$$u=x^{2} + 1$$$。

则$$$du=\left(x^{2} + 1\right)^{\prime }dx = 2 x dx$$$ (步骤见»)，并有$$$x dx = \frac{du}{2}$$$。

因此，

$${\color{red}{\int{\frac{1}{x \left(x^{2} + 1\right)} d x}}} = {\color{red}{\int{\frac{1}{2 u \left(u - 1\right)} d u}}}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u \left(u - 1\right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\frac{1}{2 u \left(u - 1\right)} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{u \left(u - 1\right)} d u}}{2}\right)}}$$

进行部分分式分解（步骤可见»）:

$$\frac{{\color{red}{\int{\frac{1}{u \left(u - 1\right)} d u}}}}{2} = \frac{{\color{red}{\int{\left(\frac{1}{u - 1} - \frac{1}{u}\right)d u}}}}{2}$$

逐项积分：

$$\frac{{\color{red}{\int{\left(\frac{1}{u - 1} - \frac{1}{u}\right)d u}}}}{2} = \frac{{\color{red}{\left(- \int{\frac{1}{u} d u} + \int{\frac{1}{u - 1} d u}\right)}}}{2}$$

设$$$v=u - 1$$$。

则$$$dv=\left(u - 1\right)^{\prime }du = 1 du$$$ (步骤见»)，并有$$$du = dv$$$。

因此，

$$- \frac{\int{\frac{1}{u} d u}}{2} + \frac{{\color{red}{\int{\frac{1}{u - 1} d u}}}}{2} = - \frac{\int{\frac{1}{u} d u}}{2} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2}$$

$$$\frac{1}{v}$$$ 的积分为 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$- \frac{\int{\frac{1}{u} d u}}{2} + \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} = - \frac{\int{\frac{1}{u} d u}}{2} + \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2}$$

回忆一下 $$$v=u - 1$$$:

$$\frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} - \frac{\int{\frac{1}{u} d u}}{2} = \frac{\ln{\left(\left|{{\color{red}{\left(u - 1\right)}}}\right| \right)}}{2} - \frac{\int{\frac{1}{u} d u}}{2}$$

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{\ln{\left(\left|{u - 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{\ln{\left(\left|{u - 1}\right| \right)}}{2} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

回忆一下 $$$u=x^{2} + 1$$$:

$$\frac{\ln{\left(\left|{-1 + {\color{red}{u}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{-1 + {\color{red}{\left(x^{2} + 1\right)}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(x^{2} + 1\right)}}}\right| \right)}}{2}$$

因此，

$$\int{\frac{1}{x \left(x^{2} + 1\right)} d x} = \frac{\ln{\left(x^{2} \right)}}{2} - \frac{\ln{\left(x^{2} + 1 \right)}}{2}$$

化简：

$$\int{\frac{1}{x \left(x^{2} + 1\right)} d x} = \ln{\left(x \right)} - \frac{\ln{\left(x^{2} + 1 \right)}}{2}$$

加上积分常数：

$$\int{\frac{1}{x \left(x^{2} + 1\right)} d x} = \ln{\left(x \right)} - \frac{\ln{\left(x^{2} + 1 \right)}}{2}+C$$

$$$\int \frac{1}{x \left(x^{2} + 1\right)}\, dx = \left(\ln\left(x\right) - \frac{\ln\left(x^{2} + 1\right)}{2}\right) + C$$$A