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

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

Integrate term by term:

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = \ln{\left(x + \sqrt{x^{2} + 1} \right)}$$$:

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

For the integral $$$\int{\ln{\left(x + \sqrt{x^{2} + 1} \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\ln{\left(x + \sqrt{x^{2} + 1} \right)}$$$ and $$$\operatorname{dv}=dx$$$.

Then $$$\operatorname{du}=\left(\ln{\left(x + \sqrt{x^{2} + 1} \right)}\right)^{\prime }dx=\frac{dx}{\sqrt{x^{2} + 1}}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d x}=x$$$ (steps can be seen »).

So,

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

Let $$$u=x^{2} + 1$$$.

Then $$$du=\left(x^{2} + 1\right)^{\prime }dx = 2 x dx$$$ (steps can be seen »), and we have that $$$x dx = \frac{du}{2}$$$.

Thus,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$:

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=- \frac{1}{2}$$$:

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

Recall that $$$u=x^{2} + 1$$$:

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

Let $$$u=x^{2} + 1$$$.

Then $$$du=\left(x^{2} + 1\right)^{\prime }dx = 2 x dx$$$ (steps can be seen »), and we have that $$$x dx = \frac{du}{2}$$$.

The integral can be rewritten as

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = \sqrt{u}$$$:

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{1}{2}$$$:

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

Recall that $$$u=x^{2} + 1$$$:

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

Therefore,

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

Add the constant of integration:

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

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