5.5 The Substitution Rule/45

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

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

${\begin{aligned}\int _{}^{}\left({\frac {x}{\sqrt[{4}]{x+2}}}\right)dx&=\int _{}^{}\left({\frac {u-2}{\sqrt[{4}]{u}}}\right)&=\int _{}^{}\left({\frac {u}{{\sqrt[{4}]{(}}u)}}-{\frac {2}{{\sqrt[{4}]{(}}u)}}\right)&=\int _{}^{}\left(u^{\frac {3}{4}}-2u^{-{\frac {1}{u}}}\right)\end{aligned}}$