$$$x \operatorname{atan}{\left(x \right)}$$$の積分

$$$\int x \operatorname{atan}{\left(x \right)}\, dx$$$ を求めよ。

積分 $$$\int{x \operatorname{atan}{\left(x \right)} d x}$$$ には、部分積分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$を用いてください。

$$$\operatorname{u}=\operatorname{atan}{\left(x \right)}$$$ と $$$\operatorname{dv}=x dx$$$ とする。

したがって、$$$\operatorname{du}=\left(\operatorname{atan}{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x^{2} + 1}$$$（手順は»を参照）および$$$\operatorname{v}=\int{x d x}=\frac{x^{2}}{2}$$$（手順は»を参照）。

したがって、

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

被積分関数を簡単化する:

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

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 $$$f{\left(x \right)} = \frac{x^{2}}{x^{2} + 1}$$$ に対して適用する:

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

分数を変形して分解する:

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

項別に積分せよ:

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

$$$c=1$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

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

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

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

したがって、

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

簡単化せよ:

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

積分定数を加える:

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

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