$$$\frac{x^{2} + 1}{\left(x^{2} + 2\right) \left(x^{2} + 3\right)}$$$の積分

$$$\int \frac{x^{2} + 1}{\left(x^{2} + 2\right) \left(x^{2} + 3\right)}\, dx$$$ を求めよ。

部分分数分解を行う (手順は»で確認できます):

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

項別に積分せよ:

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

$$$u=\frac{\sqrt{2}}{2} x$$$ とする。

すると $$$du=\left(\frac{\sqrt{2}}{2} x\right)^{\prime }dx = \frac{\sqrt{2}}{2} dx$$$（手順は»で確認できます）、$$$dx = \sqrt{2} du$$$ となります。

したがって、

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

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{\sqrt{2}}{2}$$$ と $$$f{\left(u \right)} = \frac{1}{u^{2} + 1}$$$ に対して適用する:

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

$$$\frac{1}{u^{2} + 1}$$$ の不定積分は $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$ です:

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

次のことを思い出してください $$$u=\frac{\sqrt{2}}{2} x$$$:

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

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=2$$$ と $$$f{\left(x \right)} = \frac{1}{x^{2} + 3}$$$ に対して適用する:

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

$$$u=\frac{\sqrt{3}}{3} x$$$ とする。

すると $$$du=\left(\frac{\sqrt{3}}{3} x\right)^{\prime }dx = \frac{\sqrt{3}}{3} dx$$$（手順は»で確認できます）、$$$dx = \sqrt{3} du$$$ となります。

したがって、

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

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{\sqrt{3}}{3}$$$ と $$$f{\left(u \right)} = \frac{1}{u^{2} + 1}$$$ に対して適用する:

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

$$$\frac{1}{u^{2} + 1}$$$ の不定積分は $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$ です:

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

次のことを思い出してください $$$u=\frac{\sqrt{3}}{3} x$$$:

$$- \frac{\sqrt{2} \operatorname{atan}{\left(\frac{\sqrt{2}}{2} x \right)}}{2} + \frac{2 \sqrt{3} \operatorname{atan}{\left({\color{red}{u}} \right)}}{3} = - \frac{\sqrt{2} \operatorname{atan}{\left(\frac{\sqrt{2}}{2} x \right)}}{2} + \frac{2 \sqrt{3} \operatorname{atan}{\left({\color{red}{\frac{\sqrt{3}}{3} x}} \right)}}{3}$$

したがって、

$$\int{\frac{x^{2} + 1}{\left(x^{2} + 2\right) \left(x^{2} + 3\right)} d x} = - \frac{\sqrt{2} \operatorname{atan}{\left(\frac{\sqrt{2} x}{2} \right)}}{2} + \frac{2 \sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3} x}{3} \right)}}{3}$$

積分定数を加える:

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

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