$$$\frac{x^{3}}{x^{2} - 1}$$$ 的积分

求$$$\int \frac{x^{3}}{x^{2} - 1}\, dx$$$。

由于分子次数不小于分母次数，进行多项式长除法（步骤见»）:

$${\color{red}{\int{\frac{x^{3}}{x^{2} - 1} d x}}} = {\color{red}{\int{\left(x + \frac{x}{x^{2} - 1}\right)d x}}}$$

逐项积分：

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

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：

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

设$$$u=x^{2} - 1$$$。

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

该积分可以改写为

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

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

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

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

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

回忆一下 $$$u=x^{2} - 1$$$:

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

因此，

$$\int{\frac{x^{3}}{x^{2} - 1} d x} = \frac{x^{2}}{2} + \frac{\ln{\left(\left|{x^{2} - 1}\right| \right)}}{2}$$

加上积分常数：

$$\int{\frac{x^{3}}{x^{2} - 1} d x} = \frac{x^{2}}{2} + \frac{\ln{\left(\left|{x^{2} - 1}\right| \right)}}{2}+C$$

$$$\int \frac{x^{3}}{x^{2} - 1}\, dx = \left(\frac{x^{2}}{2} + \frac{\ln\left(\left|{x^{2} - 1}\right|\right)}{2}\right) + C$$$A