What is the integral of (9x^2-3sqrt(x))/x ?

The integral of (9x^2-3sqrt(x))/x is (9x^2)/2-6sqrt(x)+C

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integral of (9x^2-3sqrt(x))/x

\int \frac{9 x ^{2} - 3 x}{x} d x

\frac{9 x ^{2}}{2} - 6 x + C

Solution steps

\int \frac{9 x ^{2} - 3 x}{x} d x

Expand

\frac{9 x ^{2} - 3 x}{x} : 9 x - \frac{3}{x}

= \int 9 x - \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 9 x d x - \int \frac{3}{x} d x

\int 9 x d x = \frac{9 x ^{2}}{2}

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

= \frac{9 x ^{2}}{2} - 6 x

Add a constant to the solution

= \frac{9 x ^{2}}{2} - 6 x + C

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What is the integral of (9x^2-3sqrt(x))/x ?
The integral of (9x^2-3sqrt(x))/x is (9x^2)/2-6sqrt(x)+C