∫ ( 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 ) ) {\displaystyle {\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}}}
![{\displaystyle \int _{}^{}\left({\frac {x}{\sqrt[{4}]{x+2}}}\right)dx}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/027085a0358008373de484ce727072a0833147c6)
![{\displaystyle {\begin{aligned}u&=x+2\\[2ex]du&=1dx\\[2ex]u-2&=x\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/7f086d72789e0c38347c5e01beecd53006c6b483)
![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/88e4aa916d6253ced41141082ede710d725be844)