How can I integrate $\ln \left( x+\sqrt{1+x^2} \right) $?
Assuming the correct question is the edited one. With integration by parts: $$\int \ln\left(x+\sqrt{1+x^{2}}\right) \,\mbox{d}x = x\ln\left(x+\sqrt{1+x^{2}}\right)-\color{blue}{\int \frac{x}{\sqrt{1+x^2}}\,\mbox{d}x}$$ Since: $$\left(\ln\left(x+\sqrt{1+x^{2}}\right) \right)' =\left( \mbox{arcsinh}\,x \right)' = \frac{1}{\sqrt{1+x^2}}$$
For the blue integral, you can substitute $u=1+x^2$ and end up with:
$$\int \ln\left(x+\sqrt{1+x^{2}}\right) \,\mbox{d}x = x\ln\left(x+\sqrt{1+x^{2}}\right)-\color{blue}{\sqrt{1+x^2}}+C$$