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