$$$\frac{x^{2}}{1 - x^{2}}$$$ 的積分

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

由於分子次數不小於分母次數，進行多項式長除法（步驟見»）:

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

逐項積分：

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

配合 $$$c=1$$$，應用常數法則 $$$\int c\, dx = c x$$$：

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

進行部分分式分解（步驟可見 »）:

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

逐項積分：

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = \frac{1}{x + 1}$$$：

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

令 $$$u=x + 1$$$。

則 $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

所以，

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

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

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

回顧一下 $$$u=x + 1$$$：

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = \frac{1}{x - 1}$$$：

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

令 $$$u=x - 1$$$。

則 $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

因此，

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

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

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

回顧一下 $$$u=x - 1$$$：

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

因此，

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

加上積分常數：

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

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