$$$\frac{1}{x^{3} \left(x + 1\right)}$$$ 的積分

求$$$\int \frac{1}{x^{3} \left(x + 1\right)}\, dx$$$。

進行部分分式分解（步驟可見 »）:

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

逐項積分：

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

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

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

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-3$$$：

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

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：

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

令 $$$u=x + 1$$$。

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

因此，

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

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

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

回顧一下 $$$u=x + 1$$$：

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

因此，

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

加上積分常數：

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

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