Answer
$2 \tan^{-1} (2x) +\dfrac{4x}{(4x^2+1)}+C$
Work Step by Step
Let us consider $x = \dfrac{1}{2} \tan \theta $
and $dx= \dfrac{1}{2} \sec^2 \theta d \theta $
Now, the given integral becomes:
$\int \dfrac{8( \dfrac{1}{2} sec^2 \theta)}{\sec ^4 \theta} d\theta=4\int cos^2 \theta d \theta $
Now, integrate:
$2(\theta+\sin \theta \cos \theta) +C=2 \tan^{-1} (2x) +\dfrac{4x}{(4x^2+1)}+C$
