## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{1}{3} (x^2-4)^{3/2}+C$
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 $\dfrac{1}{2} \int \sqrt u du=\dfrac{1}{3} (x^2-4)^{3/2}+C$