∫ 0 13 1 ( 1 + 2 x ) 2 3 d x {\displaystyle \int _{0}^{13}{\frac {1}{\sqrt[{3}]{(1+2x)^{2}}}}\,dx}
u = 1 + 2 x d u = 2 d x 1 2 d u = d x {\displaystyle {\begin{aligned}u&=1+2x\\[2ex]du&=2dx\\[2ex]{\frac {1}{2}}du&=dx\\[2ex]\end{aligned}}}
New upper limit: 27 = 1 + 2 ( 13 ) {\displaystyle 27=1+2(13)} New lower limit: 1 = 1 + 2 ( 0 ) {\displaystyle 1=1+2(0)}
∫ 0 13 1 ( 1 + 2 x ) 2 3 d x = ∫ 0 13 1 ( 1 + 2 x ) 2 3 ( d x ) = ∫ 1 27 1 u 2 3 ( 1 2 d u ) = 1 2 ∫ 1 27 u − 2 / 3 d u = 1 2 u 1 / 3 1 3 | 1 27 = 3 2 u 1 / 3 | 1 27 = 3 2 ( 27 ) 1 / 3 − 3 2 ( 1 ) 1 / 3 = 9 2 − 3 2 = 3 {\displaystyle {\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}}}