5.5 The Substitution Rule/45

$\int _{}^{}\left({\frac {x}{\sqrt[{4}]{x+2}}}\right)dx$

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

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \int_{}^{} \left(\frac {x}{\sqrt[4]{x+2}}\right)dx &=\int_{}^{} \left(\frac{u-2}{\sqrt[4]{u}}\right) \\[2ex] &=\int_{}^{} \left(\frac{u}{\sqrt[4](u)} - \frac{2}{\sqrt[4](u)}\right) \\[2ex] &=\int_{}^{} \left(u^{\frac{3}{4}} - 2u^{-\frac{1}{u}} \right) \\[2ex] &= \frac{4}{7} u^{\frac{7}{4}} - 2(\frac{4}{3})u^{\frac{3}{4} + c \\[2ex] &= \frac{4}{7} (x+2)^{\frac{7}{4}} - (\frac{8}{3})(x+2)^{\frac{3}{4} +c \\[2ex] \end{align} }