## Calculus (3rd Edition)

$\dfrac{1}{3} (4-x^2)^{3/2} -4 \sqrt {4-x^2}+C$
We will use integration by parts to obtain: $I=\int \dfrac{x^3}{\sqrt {4-x^2} } \ dx$ Let us consider the substitution $a=4-x^2 \implies da=-2x dx$ Now, $\int \dfrac{x^3}{\sqrt {4-x^2}} \ dx=\int \dfrac{(x)(x^2)}{\sqrt a} \times \dfrac{-1}{2x} \ da \\=-\dfrac{1}{2} \int \dfrac{(4-a)}{\sqrt a} \ da \\= \dfrac{1}{2} \int (a^{1/2} -4a^{-1/2} ) \ da \\= \dfrac{a^{3/2}}{3}-4a^{1/2}+C \\=\dfrac{1}{3} (4-x^2)^{3/2} -4 \sqrt {4-x^2}+C$