# How do you integrate int x^3/(x^2+2)^3 by integration by parts method?

Jan 23, 2017

Integration by parts is: $\int u \mathrm{dv} = u v - \int v \mathrm{du}$
choose $u$ so that $\mathrm{dv}$ can be integrated by variable substitution.

#### Explanation:

let $u = {x}^{2} \mathmr{and} \mathrm{dv} = \frac{x}{{x}^{2} + 2} ^ 3 \mathrm{dx}$, because this will allow:

$v = \int \frac{x}{{x}^{2} + 2} ^ 3 \mathrm{dx}$

to be integrated by letting $t = {x}^{2} + 2$, then $\mathrm{dt} = 2 x \mathrm{dx}$ or $\frac{\mathrm{dt}}{2} = x \mathrm{dx}$

This makes the integral become:

$v = \frac{1}{2} \int {t}^{-} 3 \mathrm{dt}$

$v = - \frac{1}{4} {t}^{-} 2$

Reversing the substitution:

v=-1/(4(x^2+2)^2

and $\mathrm{du} = 2 x \mathrm{dx}$

$\int {x}^{3} / {\left({x}^{2} + 2\right)}^{3} \mathrm{dx} = \left({x}^{2}\right) \left(- \frac{1}{4 {\left({x}^{2} + 2\right)}^{2}}\right) - \int - \frac{1}{{x}^{2} + 2} ^ 2 \left(2 x\right) \mathrm{dx}$

$\int {x}^{3} / {\left({x}^{2} + 2\right)}^{3} \mathrm{dx} = - {x}^{2} / \left(4 {\left({x}^{2} + 2\right)}^{2}\right) + \frac{1}{4} \int \frac{2 x}{{x}^{2} + 2} ^ 2 \mathrm{dx}$

The last integral is the same sort of variable substitution:

$\int {x}^{3} / {\left({x}^{2} + 2\right)}^{3} \mathrm{dx} = - {x}^{2} / \left(4 {\left({x}^{2} + 2\right)}^{2}\right) - \frac{1}{4 \left({x}^{2} + 2\right)} + C$