∫ 0 1 ( 3 + x x ) d x = ∫ 0 1 ( 3 + x 1 x 1 2 ) d x = ∫ 0 1 ( 3 + x 1 + 1 2 ) d x = ∫ 0 1 ( 3 + x 3 2 ) d x = 3 x + x 3 2 + 1 3 2 + 1 | 0 1 = 3 x + x 5 2 5 2 | 0 1 = 3 x + 2 x 5 2 5 | 0 1 = [ 3 ( 1 ) + 2 ( 1 ) 5 / 2 5 ] − [ 3 ( 0 ) + 2 ( 0 ) 5 / 2 5 ] = 3 + 2 5 = 15 5 + 2 5 = 17 5 {\displaystyle {\begin{aligned}\int _{0}^{1}\left(3+x{\sqrt {x}}\right)dx&=\int _{0}^{1}\left(3+x^{1}{x}^{\frac {1}{2}}\right)dx=\int _{0}^{1}\left(3+x^{1+{\frac {1}{2}}}\right)dx=\int _{0}^{1}\left(3+x^{\frac {3}{2}}\right)dx\\[2ex]&=3x+{\frac {x^{{\frac {3}{2}}+1}}{{\frac {3}{2}}+1}}{\bigg |}_{0}^{1}=3x+{\frac {x^{\tfrac {5}{2}}}{\frac {5}{2}}}{\bigg |}_{0}^{1}=3x+{\frac {2x^{\frac {5}{2}}}{5}}{\bigg |}_{0}^{1}\\[2ex]&=\left[3(1)+{\frac {2(1)^{5/2}}{5}}\right]-\left[3(0)+{\frac {2(0)^{5/2}}{5}}\right]\\[2ex]&=3+{\frac {2}{5}}={\frac {15}{5}}+{\frac {2}{5}}={\frac {17}{5}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{0}^{1}\left(3+x{\sqrt {x}}\right)dx&=\int _{0}^{1}\left(3+x^{1}{x}^{\frac {1}{2}}\right)dx=\int _{0}^{1}\left(3+x^{1+{\frac {1}{2}}}\right)dx=\int _{0}^{1}\left(3+x^{\frac {3}{2}}\right)dx\\[2ex]&=3x+{\frac {x^{{\frac {3}{2}}+1}}{{\frac {3}{2}}+1}}{\bigg |}_{0}^{1}=3x+{\frac {x^{\tfrac {5}{2}}}{\frac {5}{2}}}{\bigg |}_{0}^{1}=3x+{\frac {2x^{\frac {5}{2}}}{5}}{\bigg |}_{0}^{1}\\[2ex]&=\left[3(1)+{\frac {2(1)^{5/2}}{5}}\right]-\left[3(0)+{\frac {2(0)^{5/2}}{5}}\right]\\[2ex]&=3+{\frac {2}{5}}={\frac {15}{5}}+{\frac {2}{5}}={\frac {17}{5}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/cf00533e5237334c293ae4c07be7f695f4baa3ab)