# How do you integrate int x^3*sqrt(4+x^2)  using integration by parts?

Jan 19, 2016

not by using integration by part

$u = {x}^{2} + 4$
$\mathrm{du} = 2 x$

$\frac{1}{2} \int 2 {x}^{3} \cdot \sqrt{{x}^{2} + 4} \mathrm{du}$

$\frac{1}{2} \int {x}^{2} \cdot \sqrt{u} \mathrm{du}$

${x}^{2} = u - 4$

$\frac{1}{2} \int \left(u - 4\right) \cdot \sqrt{u} \mathrm{du}$

expand

$\frac{1}{2} \int u \cdot \sqrt{u} - 4 \sqrt{u} \mathrm{du}$

$\frac{1}{2} \int {u}^{\frac{3}{2}} - 4 \sqrt{u} \mathrm{du}$

$\frac{1}{2} \left[\frac{2}{5} {u}^{\frac{5}{2}} - \frac{8}{3} {u}^{\frac{3}{2}}\right] + C$

$\left[\frac{1}{5} {u}^{\frac{5}{2}} - \frac{4}{3} {u}^{\frac{3}{2}}\right] + C$