$\begin{aligned}
& \text { Let, } l=\int \frac{x}{\left(x^2+2 x+2\right)^2} d x \\
& l=\int \frac{x}{\left[(x+1)^2+1\right]^2} d x
\end{aligned}$
Put $x+1=\tan \theta$
$\begin{aligned}
\Rightarrow & d x=\sec ^2 \theta d \theta \\
l & =\int \frac{(\tan \theta-1) \sec ^2 \theta}{\left(\tan ^2 \theta+1\right)^2} d \theta \\
& =\int \frac{(\tan \theta-1) \sec ^2 \theta}{\sec ^4 \theta} d \theta
\end{aligned}$
$\begin{aligned}
& =\int\left(\frac{\tan \theta}{\sec ^2 \theta}-\frac{1}{\sec ^2 \theta}\right) d \theta \\
& =\int\left(\sin \theta \cos \theta-\cos ^2 \theta\right) d \theta \\
& =\frac{1}{2} \int\left(2 \sin \theta \cos \theta-2 \cos ^2 \theta\right) d \theta \\
& =\frac{1}{2} \int(\sin 2 \theta-1-\cos 2 \theta) d \theta \\
& =\frac{1}{2} \int\left(\frac{-\cos 2 \theta}{2}-\theta-\frac{\sin 2 \theta}{2}\right)+C \\
& =-\frac{1}{4}\left[\cos 2 \theta+\sin ^{2 \theta}\right]-\frac{1}{2} \theta+C \\
& =-\frac{1}{4}\left[\frac{1-\tan ^2 \theta}{1-\tan ^2 \theta}+\frac{2 \tan \theta}{1+\tan \theta}\right]-\frac{1}{2} \tan ^{-1}(x+1)+C \\
& =-\frac{1}{4}\left[\frac{1-\tan ^2 \theta+2 \tan \theta}{1+\tan ^2 \theta}\right]-\frac{1}{2} \tan ^{-1}(x+1)+C \\
& \therefore l=-\frac{1}{4}\left[\frac{x^2-2}{x^2+2 x+2}\right]-\frac{1}{2} \tan ^{-1}(x+1)+C
\end{aligned}$