What is the integral of (x^2-3)/(sqrt(x)) ?

The integral of (x^2-3)/(sqrt(x)) is 2/5 x^{5/2}-6sqrt(x)+C

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integral of (x^2-3)/(sqrt(x))

\int \frac{x ^{2} - 3}{x} d x

\frac{2}{5} x^{\frac{5}{2}} - 6 x + C

Solution steps

\int \frac{x ^{2} - 3}{x} d x

Expand

\frac{x ^{2} - 3}{x} : x^{\frac{3}{2}} - \frac{3}{x}

= \int x^{\frac{3}{2}} - \frac{3}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{\frac{3}{2}} d x - \int \frac{3}{x} d x

\int x^{\frac{3}{2}} d x = \frac{2}{5} x^{\frac{5}{2}}

\int \frac{3}{x} d x = 6 x

= \frac{2}{5} x^{\frac{5}{2}} - 6 x

Add a constant to the solution

= \frac{2}{5} x^{\frac{5}{2}} - 6 x + C

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What is the integral of (x^2-3)/(sqrt(x)) ?
The integral of (x^2-3)/(sqrt(x)) is 2/5 x^{5/2}-6sqrt(x)+C