$$$\frac{x^{2}}{x^{2} + 1}$$$の積分

$$$\int \frac{x^{2}}{x^{2} + 1}\, dx$$$ を求めよ。

分数を変形して分解する:

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

項別に積分せよ:

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

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

$$- \int{\frac{1}{x^{2} + 1} d x} + {\color{red}{\int{1 d x}}} = - \int{\frac{1}{x^{2} + 1} d x} + {\color{red}{x}}$$

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

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

したがって、

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

積分定数を加える:

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

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