∫ 1 9 x − 1 x d x = ∫ 1 9 ( x x − 1 x ) d x = ∫ 1 9 ( x x 1 / 2 − 1 x 1 / 2 ) d x = ∫ 1 9 ( x 1 / 2 − x − 1 / 2 ) d x = 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}\left({\frac {x}{\sqrt {x}}}-{\frac {1}{\sqrt {x}}}\right)dx=\int _{1}^{9}\left({\frac {x}{x^{1/2}}}-{\frac {1}{x^{1/2}}}\right)dx=\int _{1}^{9}\left(x^{1/2}-x^{-1/2}\right)dx\\[2ex]&={\frac {2x^{3/2}}{3}}-2x^{1/2}{\bigg |}_{1}^{9}\\[2ex]&=\left[{\frac {2(9)^{3/2}}{3}}-2(9)^{1/2}\right]-\left[{\frac {2(1)^{3/2}}{3}})-2(1)^{1/2}\right]\\[2ex]&=\left[{\frac {54}{3}}-6\right]-\left[{\frac {2}{3}}-2\right]={\frac {52}{3}}-4\\[2ex]&={\frac {40}{3}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{1}^{9}{\frac {x-1}{\sqrt {x}}}\,dx&=\int _{1}^{9}\left({\frac {x}{\sqrt {x}}}-{\frac {1}{\sqrt {x}}}\right)dx=\int _{1}^{9}\left({\frac {x}{x^{1/2}}}-{\frac {1}{x^{1/2}}}\right)dx=\int _{1}^{9}\left(x^{1/2}-x^{-1/2}\right)dx\\[2ex]&={\frac {2x^{3/2}}{3}}-2x^{1/2}{\bigg |}_{1}^{9}\\[2ex]&=\left[{\frac {2(9)^{3/2}}{3}}-2(9)^{1/2}\right]-\left[{\frac {2(1)^{3/2}}{3}})-2(1)^{1/2}\right]\\[2ex]&=\left[{\frac {54}{3}}-6\right]-\left[{\frac {2}{3}}-2\right]={\frac {52}{3}}-4\\[2ex]&={\frac {40}{3}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/f980f12eeb31c5e8373d1c082a8f4d2784c38d3a)