Prove

$\int \frac{1}{\sqrt{x^{2}+2 x+2}} d x=\int \frac{1}{\sqrt{(x+1)^{2}+(1)^{2}}} d x$

$\operatorname{Let} x+1=t$

$\therefore d x=d t$

$\Rightarrow \int \frac{1}{\sqrt{x^{2}+2 x+2}} d x=\int \frac{1}{\sqrt{t^{2}+1}} d t$

$=\log \left|t+\sqrt{t^{2}+1}\right|+\mathrm{C}$

$=\log \left|(x+1)+\sqrt{(x+1)^{2}+1}\right|+\mathrm{C}$

$=\log \left|(x+1)+\sqrt{x^{2}+2 x+2}\right|+\mathrm{C}$