${\displaystyle \int \frac{dx}{{\left(x+1\right)}^2\sqrt{x^2+2x+2}}}$ is equal to

${\displaystyle \int \frac{dx}{{\left(x+1\right)}^2\sqrt{x^2+2x+2}}}$

      $={\displaystyle \int \frac{dx}{{\left(x+1\right)}^2\sqrt{{\left(x+1\right)}^2+1}}={\displaystyle \int \frac{{\sec }^2\theta d\theta }{{\tan }^2\theta \sec \theta }}}$

                    $\left[Puttingx+1=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta \right]$

      $={\displaystyle \int \frac{\cos \theta }{{\sin }^2\theta }}d\theta =\frac{-1}{\sin \theta }+C=-\frac{\sqrt{{\left(x+1\right)}^2+1}}{x+1}+C$

                                         $\left[\because \tan \theta =\frac{x+1}{1}\right]$

     $=-\frac{\sqrt{x^2+2x+2}}{x+1}+C$