What is the integral of (x^3-1)/(x-2) ?

The integral of (x^3-1)/(x-2) is (x^3)/3+x^2+4x+7ln|x-2|+C

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integral of (x^3-1)/(x-2)

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

\frac{x ^{3}}{3} + x^{2} + 4 x + 7 ln ∣ x - 2 ∣ + C

Solution steps

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

Long division

\frac{x ^{3} - 1}{x - 2} : x^{2} + 2 x + 4 + \frac{7}{x - 2}

= \int x^{2} + 2 x + 4 + \frac{7}{x - 2} 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^{2} d x + \int 2 x d x + \int 4 d x + \int \frac{7}{x - 2} d x

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

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

\int 4 d x = 4 x

\int \frac{7}{x - 2} d x = 7 ln ∣ x - 2 ∣

= \frac{x ^{3}}{3} + x^{2} + 4 x + 7 ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{x ^{3}}{3} + x^{2} + 4 x + 7 ln ∣ x - 2 ∣ + C

What is the integral of (x^3-1)/(x-2) ?
The integral of (x^3-1)/(x-2) is (x^3)/3+x^2+4x+7ln|x-2|+C