5.5 The Substitution Rule/61

$\int _{0}^{13}{\frac {1}{\sqrt[{3}]{(1+2x)^{2}}}}\,dx$

${\begin{aligned}u&=1+2x\\[2ex]du&=2dx\\[2ex]{\frac {1}{2}}du&=dx\\[2ex]\end{aligned}}$

New upper limit: $27=1+2(13)$
New lower limit: $1=1+2(0)$

${\begin{aligned}\int _{0}^{13}{\frac {1}{\sqrt[{3}]{(1+2x)^{2}}}}\,dx&=\int _{0}^{13}{\frac {1}{\sqrt[{3}]{(1+2x)^{2}}}}\,(dx)\\[2ex]&=\int _{1}^{27}{\frac {1}{\sqrt[{3}]{u^{2}}}}\left({\frac {1}{2}}du\right)={\frac {1}{2}}\int _{1}^{27}{u}^{-2/3}du\\[2ex]&={\frac {1}{2}}{\frac {{u}^{1/3}}{\frac {1}{3}}}{\bigg |}_{1}^{27}={\frac {3}{2}}{u}^{1/3}{\bigg |}_{1}^{27}\\[2ex]&={\frac {3}{2}}{(27)}^{1/3}-{\frac {3}{2}}{(1)}^{1/3}\\[2ex]&={\frac {9}{2}}-{\frac {3}{2}}\\[2ex]&=3\end{aligned}}$