# How do you find the integral int_0^13dx/(root3((1+2x)^2) ?

Sep 7, 2014

By substitution,
${\int}_{0}^{13} \frac{1}{\sqrt[3]{{\left(1 + 2 x\right)}^{2}}} \mathrm{dx} = 3$

Let $u = 1 + 2 x$.
By taking the derivative,
$\frac{\mathrm{du}}{\mathrm{dx}} = 2$
By taking the reciprocal,
$\frac{\mathrm{dx}}{\mathrm{du}} = \frac{1}{2}$
By multiply by $\mathrm{du}$,
$\mathrm{dx} = \frac{\mathrm{du}}{2}$

Since $x$ goes from 0 to 13, $u$ goes from 1 to 27.
Now, we can rewrite the integral in terms of $u$.
int_0^{13}1/{root3{(1+2x)}}dx =int_1^{27}1/{root3{u^2}}{du}/2
by simplifying,
$= \frac{1}{2} {\int}_{1}^{27} {u}^{- \frac{2}{3}} \mathrm{du}$
by Power Rule,
=1/2[3u^{1/3}]_1^{27} =3/2[root3{27}-root3{1}]=3/2cdot 2=3