∫ 0 1 x ( x 3 + x 4 ) d x = ∫ 0 1 x ( x 1 3 + x 1 4 ) d x = ∫ 0 1 ( x 4 3 + x 5 4 ) d x = ( 3 x 7 3 7 + 4 x 9 4 9 ) | 0 1 = 3 7 + 4 9 = 27 + 28 7 ⋅ 9 = 55 63 {\displaystyle {\begin{aligned}\int _{0}^{1}x\left({\sqrt[{3}]{x}}+{\sqrt[{4}]{x}}\right)dx&=\int _{0}^{1}x\left(x^{\frac {1}{3}}+x^{\frac {1}{4}}\right)dx=\int _{0}^{1}\left(x^{\frac {4}{3}}+x^{\frac {5}{4}}\right)dx\\[2ex]&=\left({\frac {3x^{\frac {7}{3}}}{7}}+{\frac {4x^{\frac {9}{4}}}{9}}\right){\Bigg |}_{0}^{1}\\[2ex]&={\frac {3}{7}}+{\frac {4}{9}}={\frac {27+28}{7\cdot 9}}={\frac {55}{63}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{0}^{1}x\left({\sqrt[{3}]{x}}+{\sqrt[{4}]{x}}\right)dx&=\int _{0}^{1}x\left(x^{\frac {1}{3}}+x^{\frac {1}{4}}\right)dx=\int _{0}^{1}\left(x^{\frac {4}{3}}+x^{\frac {5}{4}}\right)dx\\[2ex]&=\left({\frac {3x^{\frac {7}{3}}}{7}}+{\frac {4x^{\frac {9}{4}}}{9}}\right){\Bigg |}_{0}^{1}\\[2ex]&={\frac {3}{7}}+{\frac {4}{9}}={\frac {27+28}{7\cdot 9}}={\frac {55}{63}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/48f8abf546e74566000f14988530adea31b106de)