$$$\frac{9}{x^{3} \left(3 x - 2\right)}$$$の積分
入力内容
$$$\int \frac{9}{x^{3} \left(3 x - 2\right)}\, dx$$$ を求めよ。
解答
定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=9$$$ と $$$f{\left(x \right)} = \frac{1}{x^{3} \left(3 x - 2\right)}$$$ に対して適用する:
$${\color{red}{\int{\frac{9}{x^{3} \left(3 x - 2\right)} d x}}} = {\color{red}{\left(9 \int{\frac{1}{x^{3} \left(3 x - 2\right)} d x}\right)}}$$
部分分数分解を行う (手順は»で確認できます):
$$9 {\color{red}{\int{\frac{1}{x^{3} \left(3 x - 2\right)} d x}}} = 9 {\color{red}{\int{\left(\frac{27}{8 \left(3 x - 2\right)} - \frac{9}{8 x} - \frac{3}{4 x^{2}} - \frac{1}{2 x^{3}}\right)d x}}}$$
項別に積分せよ:
$$9 {\color{red}{\int{\left(\frac{27}{8 \left(3 x - 2\right)} - \frac{9}{8 x} - \frac{3}{4 x^{2}} - \frac{1}{2 x^{3}}\right)d x}}} = 9 {\color{red}{\left(- \int{\frac{1}{2 x^{3}} d x} - \int{\frac{3}{4 x^{2}} d x} - \int{\frac{9}{8 x} d x} + \int{\frac{27}{8 \left(3 x - 2\right)} d x}\right)}}$$
定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{9}{8}$$$ と $$$f{\left(x \right)} = \frac{1}{x}$$$ に対して適用する:
$$- 9 \int{\frac{1}{2 x^{3}} d x} - 9 \int{\frac{3}{4 x^{2}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\int{\frac{9}{8 x} d x}}} = - 9 \int{\frac{1}{2 x^{3}} d x} - 9 \int{\frac{3}{4 x^{2}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\left(\frac{9 \int{\frac{1}{x} d x}}{8}\right)}}$$
$$$\frac{1}{x}$$$ の不定積分は $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$ です:
$$- 9 \int{\frac{1}{2 x^{3}} d x} - 9 \int{\frac{3}{4 x^{2}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{81 {\color{red}{\int{\frac{1}{x} d x}}}}{8} = - 9 \int{\frac{1}{2 x^{3}} d x} - 9 \int{\frac{3}{4 x^{2}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{81 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{8}$$
定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{3}{4}$$$ と $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$ に対して適用する:
$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\int{\frac{3}{4 x^{2}} d x}}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\left(\frac{3 \int{\frac{1}{x^{2}} d x}}{4}\right)}}$$
$$$n=-2$$$ 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:
$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\int{\frac{1}{x^{2}} d x}}}}{4}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\int{x^{-2} d x}}}}{4}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}}{4}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\left(- x^{-1}\right)}}}{4}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\left(- \frac{1}{x}\right)}}}{4}$$
定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{1}{2}$$$ と $$$f{\left(x \right)} = \frac{1}{x^{3}}$$$ に対して適用する:
$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\int{\frac{1}{2 x^{3}} d x}}} + \frac{27}{4 x} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\left(\frac{\int{\frac{1}{x^{3}} d x}}{2}\right)}} + \frac{27}{4 x}$$
$$$n=-3$$$ 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:
$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\int{\frac{1}{x^{3}} d x}}}}{2} + \frac{27}{4 x}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\int{x^{-3} d x}}}}{2} + \frac{27}{4 x}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}}{2} + \frac{27}{4 x}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\left(- \frac{x^{-2}}{2}\right)}}}{2} + \frac{27}{4 x}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}}{2} + \frac{27}{4 x}$$
定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{27}{8}$$$ と $$$f{\left(x \right)} = \frac{1}{3 x - 2}$$$ に対して適用する:
$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 {\color{red}{\int{\frac{27}{8 \left(3 x - 2\right)} d x}}} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 {\color{red}{\left(\frac{27 \int{\frac{1}{3 x - 2} d x}}{8}\right)}} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$
$$$u=3 x - 2$$$ とする。
すると $$$du=\left(3 x - 2\right)^{\prime }dx = 3 dx$$$（手順は»で確認できます）、$$$dx = \frac{du}{3}$$$ となります。
したがって、
$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{243 {\color{red}{\int{\frac{1}{3 x - 2} d x}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{243 {\color{red}{\int{\frac{1}{3 u} d u}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$
定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{1}{3}$$$ と $$$f{\left(u \right)} = \frac{1}{u}$$$ に対して適用する:
$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{243 {\color{red}{\int{\frac{1}{3 u} d u}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{243 {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{3}\right)}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$
$$$\frac{1}{u}$$$ の不定積分は $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$ です:
$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 {\color{red}{\int{\frac{1}{u} d u}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$
次のことを思い出してください $$$u=3 x - 2$$$:
$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 \ln{\left(\left|{{\color{red}{\left(3 x - 2\right)}}}\right| \right)}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$
したがって、
$$\int{\frac{9}{x^{3} \left(3 x - 2\right)} d x} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 \ln{\left(\left|{3 x - 2}\right| \right)}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$
簡単化せよ:
$$\int{\frac{9}{x^{3} \left(3 x - 2\right)} d x} = \frac{9 \left(9 x^{2} \left(- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{3 x - 2}\right| \right)}\right) + 6 x + 2\right)}{8 x^{2}}$$
積分定数を加える:
$$\int{\frac{9}{x^{3} \left(3 x - 2\right)} d x} = \frac{9 \left(9 x^{2} \left(- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{3 x - 2}\right| \right)}\right) + 6 x + 2\right)}{8 x^{2}}+C$$
解答
$$$\int \frac{9}{x^{3} \left(3 x - 2\right)}\, dx = \frac{9 \left(9 x^{2} \left(- \ln\left(\left|{x}\right|\right) + \ln\left(\left|{3 x - 2}\right|\right)\right) + 6 x + 2\right)}{8 x^{2}} + C$$$A