5.5 The Substitution Rule/45

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

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

${\begin{aligned}\int _{}^{}\left({\frac {x}{\sqrt[{4}]{x+2}}}\right)dx&=\int _{}^{}\left({\frac {u-2}{\sqrt[{4}]{u}}}\right)&=\int _{}^{}({\frac {u}{{\sqrt[{4}]{(}}u)}}-{\frac {2}{{\sqrt[{4}]{(}}u)}})\end{aligned}}$