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

The integral of (x^3)/(3x^2+3) is (x^2)/6-1/6 ln|3x^2+3|+C

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

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

\frac{x ^{2}}{6} - \frac{1}{6} ln 3 x^{2} + 3 + C

Solution steps

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

Long division

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

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

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

\int \frac{x}{3 x ^{2} + 3} d x = \frac{1}{6} ln 3 x^{2} + 3

= \frac{x ^{2}}{6} - \frac{1}{6} ln 3 x^{2} + 3

Add a constant to the solution

= \frac{x ^{2}}{6} - \frac{1}{6} ln 3 x^{2} + 3 + C

What is the integral of (x^3)/(3x^2+3) ?
The integral of (x^3)/(3x^2+3) is (x^2)/6-1/6 ln|3x^2+3|+C