$$$\frac{1}{x^{2} + 2 x + 2}$$$ 的積分

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

配方法 (步驟見 »): $$$x^{2} + 2 x + 2 = \left(x + 1\right)^{2} + 1$$$:

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

令 $$$u=x + 1$$$。

則 $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

因此，

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

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

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

回顧一下 $$$u=x + 1$$$：

$$\operatorname{atan}{\left({\color{red}{u}} \right)} = \operatorname{atan}{\left({\color{red}{\left(x + 1\right)}} \right)}$$

因此，

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

加上積分常數：

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

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