$$$\frac{f^{2}}{f^{2} + 1}$$$ 的积分

求$$$\int \frac{f^{2}}{f^{2} + 1}\, df$$$。

改写并拆分该分式:

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

逐项积分：

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

应用常数法则 $$$\int c\, df = c f$$$，使用 $$$c=1$$$：

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

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

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

因此，

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

加上积分常数：

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

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