$$$\frac{2 i n t}{x^{2} + 1}$$$ 對 $$$x$$$ 的積分

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

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

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

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

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

因此，

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

加上積分常數：

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

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