$$$\frac{3}{x^{2} + 2}$$$ 的積分

求$$$\int \frac{3}{x^{2} + 2}\, dx$$$。

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=3$$$ 與 $$$f{\left(x \right)} = \frac{1}{x^{2} + 2}$$$：

$${\color{red}{\int{\frac{3}{x^{2} + 2} d x}}} = {\color{red}{\left(3 \int{\frac{1}{x^{2} + 2} d x}\right)}}$$

令 $$$u=\frac{\sqrt{2}}{2} x$$$。

則 $$$du=\left(\frac{\sqrt{2}}{2} x\right)^{\prime }dx = \frac{\sqrt{2}}{2} dx$$$ (步驟見»)，並可得 $$$dx = \sqrt{2} du$$$。

因此，

$$3 {\color{red}{\int{\frac{1}{x^{2} + 2} d x}}} = 3 {\color{red}{\int{\frac{\sqrt{2}}{2 \left(u^{2} + 1\right)} d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{\sqrt{2}}{2}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u^{2} + 1}$$$：

$$3 {\color{red}{\int{\frac{\sqrt{2}}{2 \left(u^{2} + 1\right)} d u}}} = 3 {\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{u^{2} + 1} d u}}{2}\right)}}$$

$$$\frac{1}{u^{2} + 1}$$$ 的積分是 $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$：

$$\frac{3 \sqrt{2} {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}}{2} = \frac{3 \sqrt{2} {\color{red}{\operatorname{atan}{\left(u \right)}}}}{2}$$

回顧一下 $$$u=\frac{\sqrt{2}}{2} x$$$：

$$\frac{3 \sqrt{2} \operatorname{atan}{\left({\color{red}{u}} \right)}}{2} = \frac{3 \sqrt{2} \operatorname{atan}{\left({\color{red}{\frac{\sqrt{2}}{2} x}} \right)}}{2}$$

因此，

$$\int{\frac{3}{x^{2} + 2} d x} = \frac{3 \sqrt{2} \operatorname{atan}{\left(\frac{\sqrt{2} x}{2} \right)}}{2}$$

加上積分常數：

$$\int{\frac{3}{x^{2} + 2} d x} = \frac{3 \sqrt{2} \operatorname{atan}{\left(\frac{\sqrt{2} x}{2} \right)}}{2}+C$$

$$$\int \frac{3}{x^{2} + 2}\, dx = \frac{3 \sqrt{2} \operatorname{atan}{\left(\frac{\sqrt{2} x}{2} \right)}}{2} + C$$$A