∫ 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 | 0 π = 1 2 sin ( π ) − 1 2 sin ( 0 ) = 0 {\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}}{u}^{1/3}{\bigg |}_{0}^{\pi }\\[2ex]&={\frac {1}{2}}\sin {(\pi )}-{\frac {1}{2}}\sin {(0)}\\[2ex]&=0\end{aligned}}}
![{\displaystyle \int _{0}^{13}{\frac {1}{\sqrt[{3}]{(1+2x)^{2}}}}\,dx}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/e9f38b04c5de29c05e4c1854ebafe81175b77978)
![{\displaystyle {\begin{aligned}u&=1+2x\\[2ex]du&=2dx\\[2ex]{\frac {1}{2}}du&=dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a737fc278965117d70e906c0f6291b759f13ac66)
![{\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}}{u}^{1/3}{\bigg |}_{0}^{\pi }\\[2ex]&={\frac {1}{2}}\sin {(\pi )}-{\frac {1}{2}}\sin {(0)}\\[2ex]&=0\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/d73406fc8154f9ec1f4c135f26ef1c4ca95c2256)