Answer
$$\int _0^{13}\frac{dx}{\sqrt[3]{(1+2x)^2}}=3$$
Work Step by Step
To evaluate the integral $$\int _0^{13}\frac{dx}{\sqrt[3]{(1+2x)^2}}$$ we will use substitution $1+2x=t$ which gives us $2dx=dt\Rightarrow dx=\frac{dt}{2}$ and the integration bounds would be: for $x=0$ we have $t=1$ and for $x=13$ we have $t=27,$ so we get:
$$\int _0^{13}\frac{dx}{\sqrt[3]{(1+2x)^2}}=\int_1^{27}t^{-2/3}\frac{dt}{2}=\frac{1}{2}\left.\frac{t^{1/3}}{\frac{1}{3}}\right|_1^{27}=\frac{1}{2}\cdot3(27^{1/3}-1^{1/3})=\frac{3}{2}(3-1)=3$$
