# What is the antiderivative of x^3/(x^4-5)^3?

##### 1 Answer
Jul 8, 2016

$= - \frac{1}{8} \cdot \frac{1}{{x}^{4} - 5} ^ 2 \textcolor{g r e e n}{+ C}$

#### Explanation:

you can do this by a sub but i think that pattern spotting keeps it interesting/ enables you to do it in your head quite simply.

to start with , notice this pattern

$\frac{d}{\mathrm{dx}} \left(\frac{1}{{x}^{4} - 5} ^ 2\right) = - 2 \cdot \frac{1}{{x}^{4} - 5} ^ 3 \cdot 4 {x}^{3}$

$= - 8 \cdot {x}^{3} / {\left({x}^{4} - 5\right)}^{3}$

it's a mixture of the power rule and the chain rule.

and, so:

$\frac{d}{\mathrm{dx}} \left(\textcolor{red}{- \frac{1}{8}} \cdot \frac{1}{{x}^{4} - 5} ^ 2\right) = {x}^{3} / {\left({x}^{4} - 5\right)}^{3}$

and by the FTC we have

$\int \mathrm{dx} q \quad {x}^{3} / {\left({x}^{4} - 5\right)}^{3} = - \frac{1}{8} \cdot \frac{1}{{x}^{4} - 5} ^ 2 \textcolor{g r e e n}{+ C}$