5.5 The Substitution Rule/45

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

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

${\begin{aligned}\int _{}^{}\left({\frac {x}{\sqrt[{4}]{x+2}}}\right)dx&=\int _{}^{}\left({\frac {u-2}{\sqrt[{4}]{u}}}\right)du\\[2ex]&=\int _{}^{}\left({\frac {u}{{\sqrt[{4}]{(}}u)}}-{\frac {2}{{\sqrt[{4}]{(}}u)}}\right)du\\[2ex]&=\int _{}^{}\left(u^{\frac {3}{4}}-2u^{-{\frac {1}{u}}}\right)du\\[2ex]&={\frac {4}{7}}u^{\frac {7}{4}}-2({\frac {4}{3}})u^{\frac {3}{4}}+c\\[2ex]&={\frac {4}{7}}(x+2)^{\frac {7}{4}}-{\frac {8}{3}}(x+2)^{\frac {3}{4}}+c\\[2ex]\end{aligned}}$