$$$\frac{x^{2}}{x^{2} + 1}$$$ 的积分

求$$$\int \frac{x^{2}}{x^{2} + 1}\, dx$$$。

改写并拆分该分式:

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

逐项积分：

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

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：

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

$$$\frac{1}{x^{2} + 1}$$$ 的积分为 $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$:

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

因此，

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

加上积分常数：

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

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