∫ ( 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 ) d u = ∫ ( u ( 4 u ) − 2 ( 4 u ) ) d u = ∫ ( u 3 4 − 2 u − 1 u ) d 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)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}}}
![{\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)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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/0e567d41387d432138fb7c8264f06d35663fac70)