$$$\ln\left(x^{2} + 1\right)$$$ 的积分

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

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

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

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

因此，

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

对 $$$c=2$$$ 和 $$$f{\left(x \right)} = \frac{x^{2}}{x^{2} + 1}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

改写并拆分该分式:

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

逐项积分：

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

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：

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

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

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

因此，

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

加上积分常数：

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

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