Answer
$\dfrac{(\sqrt {x^2+4})^3}{3}-4 \sqrt {x^2+4}+C$
Work Step by Step
Let us consider $x = 2 \tan \theta $
and $dx=2 \sec^2 \theta d\theta $
Now, the given integral becomes:
$\int \dfrac{8 \tan^3 \theta (2 \sec^2 \theta) }{\sqrt{4 \tan^2 \theta+4}} d \theta =\int \dfrac{1 6\tan^3 \theta (\sec^2 \theta) }{2 \sqrt{(\tan^2 \theta+1})} d \theta =8 \int (\sec^2 d \theta -1) (\sec \theta\tan \theta) d \theta$
Let us suppose $u=\sec \theta \implies du =\sec \theta \theta d\theta $
Thus, we have $8 \int (\sec^2 d \theta -1) (\sec \theta\tan \theta) d \theta=\int 8(u^2-1) du$
Now, integrate and plug in $u=\sec \theta$
$8( \dfrac{u^3}{3}- u)+C=8( \dfrac{\sec^3 \theta}{3}-\sec \theta)+C$
Now, plug in $x = 2 \tan \theta \implies \dfrac{\sin \theta}{\cos \theta} =(x/2)$ and $\sec \theta =(\sqrt{x^2+4})/2 $:
Thus, $8( \dfrac{\sec^3 \theta}{3}-\sec \theta)+C=8[\dfrac{(\dfrac{\sqrt {x^2+4}}{2})^3}{3})]-8 \dfrac{\sqrt {x^2+4}}{2}+C$
or, $=\dfrac{(\sqrt {x^2+4})^3}{3}-4 \sqrt {x^2+4}+C$
