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

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

逐项积分：

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

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

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

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

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

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

该积分可以改写为

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

设$$$u=x^{2} + 1$$$。

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

该积分可以改写为

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

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=- \frac{1}{2}$$$：

$$\frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} + \int{\frac{x \sqrt{x^{2} + 1}}{2} d x} - \frac{{\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}}{4}=\frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} + \int{\frac{x \sqrt{x^{2} + 1}}{2} d x} - \frac{{\color{red}{\int{u^{- \frac{1}{2}} d u}}}}{4}=\frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} + \int{\frac{x \sqrt{x^{2} + 1}}{2} d x} - \frac{{\color{red}{\frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}}{4}=\frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} + \int{\frac{x \sqrt{x^{2} + 1}}{2} d x} - \frac{{\color{red}{\left(2 u^{\frac{1}{2}}\right)}}}{4}=\frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} + \int{\frac{x \sqrt{x^{2} + 1}}{2} d x} - \frac{{\color{red}{\left(2 \sqrt{u}\right)}}}{4}$$

回忆一下 $$$u=x^{2} + 1$$$:

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

设$$$u=x^{2} + 1$$$。

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

所以，

$$\frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} - \frac{\sqrt{x^{2} + 1}}{2} + {\color{red}{\int{\frac{x \sqrt{x^{2} + 1}}{2} d x}}} = \frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} - \frac{\sqrt{x^{2} + 1}}{2} + {\color{red}{\int{\frac{\sqrt{u}}{4} d u}}}$$

对 $$$c=\frac{1}{4}$$$ 和 $$$f{\left(u \right)} = \sqrt{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$\frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} - \frac{\sqrt{x^{2} + 1}}{2} + {\color{red}{\int{\frac{\sqrt{u}}{4} d u}}} = \frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} - \frac{\sqrt{x^{2} + 1}}{2} + {\color{red}{\left(\frac{\int{\sqrt{u} d u}}{4}\right)}}$$

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=\frac{1}{2}$$$：

$$\frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} - \frac{\sqrt{x^{2} + 1}}{2} + \frac{{\color{red}{\int{\sqrt{u} d u}}}}{4}=\frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} - \frac{\sqrt{x^{2} + 1}}{2} + \frac{{\color{red}{\int{u^{\frac{1}{2}} d u}}}}{4}=\frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} - \frac{\sqrt{x^{2} + 1}}{2} + \frac{{\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}}{4}=\frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} - \frac{\sqrt{x^{2} + 1}}{2} + \frac{{\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}}{4}$$

回忆一下 $$$u=x^{2} + 1$$$:

$$\frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} - \frac{\sqrt{x^{2} + 1}}{2} + \frac{{\color{red}{u}}^{\frac{3}{2}}}{6} = \frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} - \frac{\sqrt{x^{2} + 1}}{2} + \frac{{\color{red}{\left(x^{2} + 1\right)}}^{\frac{3}{2}}}{6}$$

因此，

$$\int{\left(\frac{x \sqrt{x^{2} + 1}}{2} + \frac{\ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2}\right)d x} = \frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} + \frac{\left(x^{2} + 1\right)^{\frac{3}{2}}}{6} - \frac{\sqrt{x^{2} + 1}}{2}$$

加上积分常数：

$$\int{\left(\frac{x \sqrt{x^{2} + 1}}{2} + \frac{\ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2}\right)d x} = \frac{x \ln{\left(x + \sqrt{x^{2} + 1} \right)}}{2} + \frac{\left(x^{2} + 1\right)^{\frac{3}{2}}}{6} - \frac{\sqrt{x^{2} + 1}}{2}+C$$

$$$\int \left(\frac{x \sqrt{x^{2} + 1}}{2} + \frac{\ln\left(x + \sqrt{x^{2} + 1}\right)}{2}\right)\, dx = \left(\frac{x \ln\left(x + \sqrt{x^{2} + 1}\right)}{2} + \frac{\left(x^{2} + 1\right)^{\frac{3}{2}}}{6} - \frac{\sqrt{x^{2} + 1}}{2}\right) + C$$$A