$$$\operatorname{atan}{\left(x \right)}$$$ 的積分
您的輸入
求$$$\int \operatorname{atan}{\left(x \right)}\, dx$$$。
解答
對於積分 $$$\int{\operatorname{atan}{\left(x \right)} d x}$$$,使用分部積分法 $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。
令 $$$\operatorname{u}=\operatorname{atan}{\left(x \right)}$$$ 與 $$$\operatorname{dv}=dx$$$。
則 $$$\operatorname{du}=\left(\operatorname{atan}{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x^{2} + 1}$$$(步驟見 »),且 $$$\operatorname{v}=\int{1 d x}=x$$$(步驟見 »)。
因此,
$${\color{red}{\int{\operatorname{atan}{\left(x \right)} d x}}}={\color{red}{\left(\operatorname{atan}{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x^{2} + 1} d x}\right)}}={\color{red}{\left(x \operatorname{atan}{\left(x \right)} - \int{\frac{x}{x^{2} + 1} d x}\right)}}$$
令 $$$u=x^{2} + 1$$$。
則 $$$du=\left(x^{2} + 1\right)^{\prime }dx = 2 x dx$$$ (步驟見»),並可得 $$$x dx = \frac{du}{2}$$$。
該積分可改寫為
$$x \operatorname{atan}{\left(x \right)} - {\color{red}{\int{\frac{x}{x^{2} + 1} d x}}} = x \operatorname{atan}{\left(x \right)} - {\color{red}{\int{\frac{1}{2 u} d u}}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u}$$$:
$$x \operatorname{atan}{\left(x \right)} - {\color{red}{\int{\frac{1}{2 u} d u}}} = x \operatorname{atan}{\left(x \right)} - {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}}$$
$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$x \operatorname{atan}{\left(x \right)} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = x \operatorname{atan}{\left(x \right)} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$
回顧一下 $$$u=x^{2} + 1$$$:
$$x \operatorname{atan}{\left(x \right)} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = x \operatorname{atan}{\left(x \right)} - \frac{\ln{\left(\left|{{\color{red}{\left(x^{2} + 1\right)}}}\right| \right)}}{2}$$
因此,
$$\int{\operatorname{atan}{\left(x \right)} d x} = x \operatorname{atan}{\left(x \right)} - \frac{\ln{\left(x^{2} + 1 \right)}}{2}$$
加上積分常數:
$$\int{\operatorname{atan}{\left(x \right)} d x} = x \operatorname{atan}{\left(x \right)} - \frac{\ln{\left(x^{2} + 1 \right)}}{2}+C$$
答案
$$$\int \operatorname{atan}{\left(x \right)}\, dx = \left(x \operatorname{atan}{\left(x \right)} - \frac{\ln\left(x^{2} + 1\right)}{2}\right) + C$$$A