Answer
$\int_{0}^{13}\frac{dx}{\sqrt[3] {(1+2x)^2}} = 3$
Work Step by Step
$\int_{0}^{13}\frac{dx}{\sqrt[3] {(1+2x)^2}}$
Let $u = 1+2x$
$\frac{du}{dx} = 2$
$dx = \frac{du}{2}$
When $x = 0$, then $u = 1$
When $x = 13$, then $u = 27$
$\int_{1}^{27} \frac{1}{2}\cdot \frac{du}{\sqrt[3] {u)^2}}$
$=\int_{1}^{27} \frac{1}{2}\cdot u^{-2/3}~du$
$= \frac{1}{2}~(3u^{1/3}~\vert_{1}^{27})$
$= \frac{3}{2}~(\sqrt[3] u~\vert_{1}^{27})$
$= \frac{3}{2}~(\sqrt[3] 27-\sqrt[3] 1)$
$= \frac{3}{2}~(3-1)$
$= 3$
