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

The integral of (x^2-9)/x is (x^2)/2-9ln|x|+C

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integral of (x^2-9)/x

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

\frac{x ^{2}}{2} - 9 ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

\int \frac{9}{x} d x = 9 ln ∣ x ∣

= \frac{x ^{2}}{2} - 9 ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - 9 ln ∣ x ∣ + C

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What is the integral of (x^2-9)/x ?
The integral of (x^2-9)/x is (x^2)/2-9ln|x|+C