Question
Integrate the function $\frac{1}{\sqrt{x^{2}+2 x+2}}$ 

Answer

Clearly, $\int \frac{1}{\sqrt{x^{2}+2 x+2}} d x=\int \frac{1}{\sqrt{(x+1)^{2}+(1)^{2}}} d x$ 
Let x + 1 = t
$\Rightarrow$ dx = dt
$\Rightarrow \int \frac{1}{\sqrt{(x+1)^{2}+(1)^{2}}} d x=\int \frac{1}{\sqrt{t^{2}+1}} d t$ 
$=\log |t+\sqrt{t^{2}+1}|+C$ 
$=\log |(x+1)+\sqrt{(x+1)^{2}+1}|+C$ 
$=\log |(x+1)+\sqrt{x^{2}+2 x+2}|+C$

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