$\int \frac{dx}{x^2+2x+2}$ is equal to:

Let $A=\int \frac{dx}{x^2+2x+2}=\int \frac{dx}{(1+x{)}^2+1}$

Put $1+x=\tan \theta$  ………,.. (i)

and $\mathrm{d}x={\sec }^2\theta d\theta$ Thus $\mathrm{A}=\int \frac{{\sec }^2\theta \mathrm{d}\theta }{1+{\tan }^2\theta }=\int \frac{{\sec }^2\theta }{{\sec }^2\theta }\cdot \mathrm{d}\theta =\theta +\mathrm{c}$

From (i)   

              $\theta ={\tan }^{-1}(1+x)$

$\therefore \mathrm{A}={\tan }^{-1}(1+\mathrm{x})+\mathrm{c}$