5.5 The Substitution Rule/61

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

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

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: 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 1 = 1+2(0)}

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