∫ 1 9 x − 1 x d x = ∫ 1 9 x x − 1 x = ∫ 1 9 x x 1 / 2 − 1 x 1 / 2 = ∫ 1 9 x 1 / 2 − x − 1 / 2 = ∫ 1 9 2 x 3 / 2 3 − 2 x 1 / 2 | 1 9 = ( 2 ( 9 ) 3 / 2 3 ) − 2 ( 9 ) 1 / 2 ) − ( 2 ( 1 ) 3 / 2 3 ) − 2 ( 1 ) 1 / 2 ) = ( 54 3 − 6 ) − ( 2 3 − 2 ) = 52 3 − 4 = 40 3 {\displaystyle {\begin{aligned}\int _{1}^{9}{\frac {x-1}{\sqrt {x}}}dx=\int _{1}^{9}{\frac {x}{\sqrt {x}}}-{\frac {1}{\sqrt {x}}}=\int _{1}^{9}{\frac {x}{x^{1/2}}}-{\frac {1}{x^{1/2}}}=\int _{1}^{9}x^{1/2}-x^{-1/2}=\int _{1}^{9}{\frac {2x^{3/2}}{3}}-2x^{1/2}{\bigg |}_{1}^{9}\\[2ex]&=({\frac {2(9)^{3/2}}{3}})-2(9)^{1/2})-({\frac {2(1)^{3/2}}{3}})-2(1)^{1/2})=({\frac {54}{3}}-6)-({\frac {2}{3}}-2)&={\frac {52}{3}}-4={\frac {40}{3}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{1}^{9}{\frac {x-1}{\sqrt {x}}}dx=\int _{1}^{9}{\frac {x}{\sqrt {x}}}-{\frac {1}{\sqrt {x}}}=\int _{1}^{9}{\frac {x}{x^{1/2}}}-{\frac {1}{x^{1/2}}}=\int _{1}^{9}x^{1/2}-x^{-1/2}=\int _{1}^{9}{\frac {2x^{3/2}}{3}}-2x^{1/2}{\bigg |}_{1}^{9}\\[2ex]&=({\frac {2(9)^{3/2}}{3}})-2(9)^{1/2})-({\frac {2(1)^{3/2}}{3}})-2(1)^{1/2})=({\frac {54}{3}}-6)-({\frac {2}{3}}-2)&={\frac {52}{3}}-4={\frac {40}{3}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/36d96556237243d80bd469d511066428185f2408)