# What is the antiderivative of (2x-1)^3?

Aug 6, 2015

$\int {\left(2 x - 1\right)}^{2} \mathrm{dx} = \frac{1}{8} {\left(2 x - 1\right)}^{4} + c$

#### Explanation:

To solve:
$\int {\left(2 x - 1\right)}^{2} \mathrm{dx}$

We use the substitution:
$u = 2 x - 1$ => $\frac{\mathrm{du}}{\mathrm{dx}} = 2$ => $\mathrm{dx} = \frac{1}{2} \mathrm{du}$

This gives:
$\int {\left(2 x - 1\right)}^{3} \mathrm{dx} = \frac{1}{2} \int {u}^{3} \mathrm{du} = \frac{1}{8} {u}^{4} + c$

Transforming back to the original variable we have:
$\frac{1}{8} {u}^{4} + c = \frac{1}{8} {\left(2 x - 1\right)}^{4} + c$