$$$\frac{1}{x^{2} + 4}$$$ 的积分

求$$$\int \frac{1}{x^{2} + 4}\, dx$$$。

设$$$u=\frac{x}{2}$$$。

则$$$du=\left(\frac{x}{2}\right)^{\prime }dx = \frac{dx}{2}$$$ (步骤见»)，并有$$$dx = 2 du$$$。

因此，

$${\color{red}{\int{\frac{1}{x^{2} + 4} d x}}} = {\color{red}{\int{\frac{1}{2 \left(u^{2} + 1\right)} d u}}}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u^{2} + 1}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

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

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

回忆一下 $$$u=\frac{x}{2}$$$:

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

因此，

$$\int{\frac{1}{x^{2} + 4} d x} = \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{2}$$

加上积分常数：

$$\int{\frac{1}{x^{2} + 4} d x} = \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{2}+C$$

$$$\int \frac{1}{x^{2} + 4}\, dx = \frac{\operatorname{atan}{\left(\frac{x}{2} \right)}}{2} + C$$$A