How do you integrate int (x^2-1)/(x^(3/2))dx?

$\int \frac{{x}^{2} - 1}{{x}^{\frac{3}{2}}} \mathrm{dx} = \frac{2}{3} {x}^{\frac{3}{2}} + 2 {x}^{- \frac{1}{2}} + C$
The integral is solved immediately if we make the previous division of the numerator between the denominator. To do this, we decompose the division into two terms and then simplify the powers of $x$:
$\int \frac{{x}^{2} - 1}{{x}^{\frac{3}{2}}} \mathrm{dx} = \int \left({x}^{2} / {x}^{\frac{3}{2}} - \frac{1}{x} ^ \left\{\frac{3}{2}\right\}\right) \mathrm{dx} = \int {x}^{\frac{1}{2}} \mathrm{dx} - \int {x}^{- \frac{3}{2}} \mathrm{dx} =$
$= \frac{{x}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - \frac{{x}^{- \frac{3}{2} + 1}}{- \frac{3}{2} + 1} + C = \frac{2}{3} {x}^{\frac{3}{2}} + 2 {x}^{- \frac{1}{2}} + C$.