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

(N/A) To integrate $\int \frac{1}{\sqrt{x^{2}+2x+2}} dx$,we first complete the square for the quadratic expression in the denominator.
$x^{2}+2x+2 = (x^{2}+2x+1)+1 = (x+1)^{2}+1^{2}$.
Now,the integral becomes $\int \frac{1}{\sqrt{(x+1)^{2}+1^{2}}} dx$.
Let $t = x+1$,then $dt = dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{t^{2}+a^{2}}} dt = \log |t + \sqrt{t^{2}+a^{2}}| + C$,where $a=1$:
$\int \frac{1}{\sqrt{t^{2}+1^{2}}} dt = \log |t + \sqrt{t^{2}+1}| + C$.
Substituting $t = x+1$ back into the expression:
$= \log |(x+1) + \sqrt{(x+1)^{2}+1}| + C$
$= \log |(x+1) + \sqrt{x^{2}+2x+2}| + C$,where $C$ is an arbitrary constant.