# How do you find the integral intx^3/(sqrt(x^2+9))dx ?

Aug 29, 2014

$= {\left({x}^{2} + 9\right)}^{\frac{3}{2}} / 3 - 9 \sqrt{{x}^{2} + 9} + c$, where $c$ is a constant

Explanation :

$= \int {x}^{3} / \sqrt{{x}^{2} + 9} \mathrm{dx}$

Let's assume ${x}^{2} + 9 = {t}^{2}$, then

$2 x \mathrm{dx} = 2 t \mathrm{dt}$, $\implies x \mathrm{dx} = t \mathrm{dt}$

$= \int \frac{\left({t}^{2} - 9\right) t}{t} \mathrm{dt}$

$= \int {t}^{2} \mathrm{dt} - \int 9 \mathrm{dt}$

$= {t}^{3} / 3 - 9 t + c$, where $c$ is a constant

Substituting $t$ back,

$= {\left({x}^{2} + 9\right)}^{\frac{3}{2}} / 3 - 9 \sqrt{{x}^{2} + 9} + c$, where $c$ is a constant