What is the integral of (x^2)/(3x+18) ?

The integral of (x^2)/(3x+18) is (x^2)/6-2x+12ln|x+6|+C

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integral of (x^2)/(3x+18)

\int \frac{x ^{2}}{3 x + 18} d x

\frac{x ^{2}}{6} - 2 x + 12 ln ∣ x + 6 ∣ + C

Solution steps

\int \frac{x ^{2}}{3 x + 18} d x

Long division

\frac{x ^{2}}{3 x + 18} : \frac{x}{3} - 2 + \frac{12}{x + 6}

= \int \frac{x}{3} - 2 + \frac{12}{x + 6} 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 \frac{x}{3} d x - \int 2 d x + \int \frac{12}{x + 6} d x

\int \frac{x}{3} d x = \frac{x ^{2}}{6}

\int 2 d x = 2 x

\int \frac{12}{x + 6} d x = 12 ln ∣ x + 6 ∣

= \frac{x ^{2}}{6} - 2 x + 12 ln ∣ x + 6 ∣

Add a constant to the solution

= \frac{x ^{2}}{6} - 2 x + 12 ln ∣ x + 6 ∣ + C

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What is the integral of (x^2)/(3x+18) ?
The integral of (x^2)/(3x+18) is (x^2)/6-2x+12ln|x+6|+C