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

The integral of (x^3-2)/(3x-3) is (x^3)/9+(x^2)/6+1/3 x-1/3 ln|3x-3|+C

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

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

\frac{x ^{3}}{9} + \frac{x ^{2}}{6} + \frac{1}{3} x - \frac{1}{3} ln ∣ 3 x - 3 ∣ + C

Solution steps

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

Long division

\frac{x ^{3} - 2}{3 x - 3} : \frac{x ^{2}}{3} + \frac{x}{3} + \frac{1}{3} - \frac{1}{3 x - 3}

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

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

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

\int \frac{1}{3} d x = \frac{1}{3} x

\int \frac{1}{3 x - 3} d x = \frac{1}{3} ln ∣ 3 x - 3 ∣

= \frac{x ^{3}}{9} + \frac{x ^{2}}{6} + \frac{1}{3} x - \frac{1}{3} ln ∣ 3 x - 3 ∣

Add a constant to the solution

= \frac{x ^{3}}{9} + \frac{x ^{2}}{6} + \frac{1}{3} x - \frac{1}{3} ln ∣ 3 x - 3 ∣ + C

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What is the integral of (x^3-2)/(3x-3) ?
The integral of (x^3-2)/(3x-3) is (x^3)/9+(x^2)/6+1/3 x-1/3 ln|3x-3|+C