# What is the antiderivative of x/(7x^2+3)^5?

May 29, 2016

$- \frac{1}{56 {\left(7 {x}^{2} + 3\right)}^{4}} + C$

#### Explanation:

We wish to find:

$\int \frac{x}{7 {x}^{2} + 3} ^ 5 \mathrm{dx}$

This is a prime case for substitution. Let $u = 7 {x}^{2} + 3$ so that $\mathrm{du} = 14 x \mathrm{dx}$.

Note that we already have $x \mathrm{dx}$ in the numerator, so we need to multiply the integrand by $14$ to obtain $\mathrm{du}$. Balance this by multiplying the exterior of the integral by $\frac{1}{14}$.

$= \frac{1}{14} \int \frac{14 x}{7 {x}^{2} + 3} ^ 5 \mathrm{dx}$

Now substitute in $u$ and $\mathrm{du}$.

$= \frac{1}{14} \int \frac{\mathrm{du}}{u} ^ 5 = \frac{1}{14} \int {u}^{-} 5 \mathrm{du}$

Integrating ${u}^{-} 5$ results in ${u}^{- 5 + 1} / \left(- 5 + 1\right) + C$:

$= \frac{1}{14} \left({u}^{-} \frac{4}{- 4}\right) + C = - \frac{1}{56 {u}^{4}} + C = - \frac{1}{56 {\left(7 {x}^{2} + 3\right)}^{4}} + C$