Answer
$\dfrac{1}{3} (x^2-4)^{3/2}+C$
Work Step by Step
Let us consider $u=x^2-4 \implies du =2x dx$
Now, the given integral can be written as:
$=\dfrac{1}{2} \int 2x \sqrt{x^2-4} dx$
and $=\dfrac{1}{2} \int \sqrt u du$
Now, plug in $u=x^2-4$
Thus, we have
$\dfrac{1}{2} \int \sqrt u du=\dfrac{1}{3} (x^2-4)^{3/2}+C$
