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$$$\operatorname{atan}{\left(x \right)}$$$ 的积分
您的输入
求$$$\int \operatorname{atan}{\left(x \right)}\, dx$$$。
解答
对于积分$$$\int{\operatorname{atan}{\left(x \right)} d x}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。
设 $$$\operatorname{u}=\operatorname{atan}{\left(x \right)}$$$ 和 $$$\operatorname{dv}=dx$$$。
则 $$$\operatorname{du}=\left(\operatorname{atan}{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x^{2} + 1}$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{1 d x}=x$$$ (步骤见 »)。
因此，
$${\color{red}{\int{\operatorname{atan}{\left(x \right)} d x}}}={\color{red}{\left(\operatorname{atan}{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x^{2} + 1} d x}\right)}}={\color{red}{\left(x \operatorname{atan}{\left(x \right)} - \int{\frac{x}{x^{2} + 1} d x}\right)}}$$
设$$$u=x^{2} + 1$$$。
则$$$du=\left(x^{2} + 1\right)^{\prime }dx = 2 x dx$$$ (步骤见»)，并有$$$x dx = \frac{du}{2}$$$。
因此，
$$x \operatorname{atan}{\left(x \right)} - {\color{red}{\int{\frac{x}{x^{2} + 1} d x}}} = x \operatorname{atan}{\left(x \right)} - {\color{red}{\int{\frac{1}{2 u} d u}}}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：
$$x \operatorname{atan}{\left(x \right)} - {\color{red}{\int{\frac{1}{2 u} d u}}} = x \operatorname{atan}{\left(x \right)} - {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}}$$
$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$x \operatorname{atan}{\left(x \right)} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = x \operatorname{atan}{\left(x \right)} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$
回忆一下 $$$u=x^{2} + 1$$$:
$$x \operatorname{atan}{\left(x \right)} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = x \operatorname{atan}{\left(x \right)} - \frac{\ln{\left(\left|{{\color{red}{\left(x^{2} + 1\right)}}}\right| \right)}}{2}$$
因此，
$$\int{\operatorname{atan}{\left(x \right)} d x} = x \operatorname{atan}{\left(x \right)} - \frac{\ln{\left(x^{2} + 1 \right)}}{2}$$
加上积分常数：
$$\int{\operatorname{atan}{\left(x \right)} d x} = x \operatorname{atan}{\left(x \right)} - \frac{\ln{\left(x^{2} + 1 \right)}}{2}+C$$
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