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: $1=1+2(0)$

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} \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{align} }