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

Aug 29, 2014

$= \frac{1}{2} \cdot \frac{x}{\sqrt{2 + {x}^{2}}} + c$, where $c$ is a constant

Explanation :

$= \int {x}^{2} / \left({\left({x}^{2} + 2\right)}^{\frac{3}{2}}\right) \mathrm{dx}$

$= \int {x}^{2} / \left({x}^{3} {\left(1 + \frac{2}{x} ^ 2\right)}^{\frac{3}{2}}\right) \mathrm{dx}$

$= \int \frac{1}{x {\left(1 + \frac{2}{x} ^ 2\right)}^{\frac{3}{2}}} \mathrm{dx}$

let's assume $\frac{2}{x} ^ 2 = t$

then $- \frac{4}{x} \mathrm{dx} = \mathrm{dt}$

$= \int - \frac{\mathrm{dt}}{4 {\left(1 + t\right)}^{\frac{3}{2}}}$

$= - \frac{1}{4} \int {\left(1 + t\right)}^{- \frac{3}{2}} \mathrm{dt}$

$= - \frac{1}{4} \cdot \frac{{\left(1 + t\right)}^{- \frac{1}{2}}}{- \frac{1}{2}}$

$= \frac{1}{2} \cdot \frac{1}{\sqrt{1 + t}} + c$, where $c$ is a constant

Substituting $t$ back,

$= \frac{1}{2} \cdot \frac{1}{\sqrt{1 + \frac{2}{x} ^ 2}} + c$, where $c$ is a constant

$= \frac{1}{2} \cdot \frac{x}{\sqrt{2 + {x}^{2}}} + c$, where $c$ is a constant