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

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

部分分数分解を行う (手順は»で確認できます):

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

項別に積分せよ:

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

$$$\frac{1}{x}$$$ の不定積分は $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$ です:

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

$$$n=-3$$$ を用いて、べき乗の法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

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

$$$n=-2$$$ を用いて、べき乗の法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

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

$$$u=x + 1$$$ とする。

すると $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$（手順は»で確認できます）、$$$dx = du$$$ となります。

積分は次のようになります

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

$$$\frac{1}{u}$$$ の不定積分は $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$ です:

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

次のことを思い出してください $$$u=x + 1$$$:

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

したがって、

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

積分定数を加える:

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

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