$$$\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}$$$ 的积分

求$$$\int \operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}\, dx$$$。

对于积分$$$\int{\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)} d x}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

设 $$$\operatorname{u}=\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}$$$ 和 $$$\operatorname{dv}=dx$$$。

则 $$$\operatorname{du}=\left(\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)}\right)^{\prime }dx=\frac{2 \left(1 - x^{2}\right)}{\left(x^{2} + 1\right) \sqrt{x^{4} - 2 x^{2} + 1}} dx$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{1 d x}=x$$$ (步骤见 »)。

因此，

$${\color{red}{\int{\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)} d x}}}={\color{red}{\left(\operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)} \cdot x-\int{x \cdot \frac{2 \left(1 - x^{2}\right)}{\left(x^{2} + 1\right) \sqrt{x^{4} - 2 x^{2} + 1}} d x}\right)}}={\color{red}{\left(x \operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)} - \int{\left(- \frac{2 x \left(x - 1\right) \left(x + 1\right)}{\left(x^{2} + 1\right) \left|{x - 1}\right| \left|{x + 1}\right|}\right)d x}\right)}}$$

设$$$u=x^{2} + 1$$$。

则$$$du=\left(x^{2} + 1\right)^{\prime }dx = 2 x dx$$$ (步骤见»)，并有$$$x dx = \frac{du}{2}$$$。

积分变为

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

化简被积函数:

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

对 $$$c=-1$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

回忆一下 $$$u=x^{2} + 1$$$:

$$x \operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)} + \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = x \operatorname{asin}{\left(\frac{2 x}{x^{2} + 1} \right)} + \ln{\left(\left|{{\color{red}{\left(x^{2} + 1\right)}}}\right| \right)}$$

因此，

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

加上积分常数：

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

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