5.5 The Substitution Rule/61

$\int _{0}^{13}{\frac {dx}{\sqrt[{3}]{(1+2x)^{2}}}}$ = $\int _{0}^{13}{\frac {1}{2^{3}{\sqrt {t^{2}}}}}dt$ = ${\frac {1}{2}}\int _{0}^{13}{\frac {1}{\sqrt[{3}]{t^{2}}}}dt$ = ${\frac {1}{2}}\int _{0}^{13}{\frac {1}{t^{\frac {2}{3}}}}dt$ = ${\frac {1}{2}}3{\sqrt[{3}]{t}}$ = ${\frac {1}{2}}3{\sqrt[{3}]{1+2x}}$ = ${\frac {3}{2}}{\sqrt[{3}]{1+2x}}$ = ${\frac {3}{2}}{\sqrt[{3}]{1+2x}}{\bigg |}_{0}^{13}$ = ${\frac {3}{2}}{\sqrt[{3}]{1+2x*13}}-{\frac {3}{2}}{\sqrt[{3}]{1+2*0}}$ = $\ 3$ \\[2ex]

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

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

New upper limit: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \27 = 1+2(13)}
New lower limit: $1=1+2(0)$

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