What is the integral of (x^5)/(x-2) ?

The integral of (x^5)/(x-2) is (x^5)/5+(x^4)/2+(4x^3)/3+4x^2+16x+32ln|x-2|+C

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integral of (x^5)/(x-2)

\int \frac{x ^{5}}{x - 2} d x

\frac{x ^{5}}{5} + \frac{x ^{4}}{2} + \frac{4 x ^{3}}{3} + 4 x^{2} + 16 x + 32 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x ^{5}}{x - 2} d x

Long division

\frac{x ^{5}}{x - 2} : x^{4} + 2 x^{3} + 4 x^{2} + 8 x + 16 + \frac{32}{x - 2}

= \int x^{4} + 2 x^{3} + 4 x^{2} + 8 x + 16 + \frac{32}{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^{4} d x + \int 2 x^{3} d x + \int 4 x^{2} d x + \int 8 x d x + \int 16 d x + \int \frac{32}{x - 2} d x

\int x^{4} d x = \frac{x ^{5}}{5}

\int 2 x^{3} d x = \frac{x ^{4}}{2}

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 8 x d x = 4 x^{2}

\int 16 d x = 16 x

\int \frac{32}{x - 2} d x = 32 ln ∣ x - 2 ∣

= \frac{x ^{5}}{5} + \frac{x ^{4}}{2} + \frac{4 x ^{3}}{3} + 4 x^{2} + 16 x + 32 ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{x ^{5}}{5} + \frac{x ^{4}}{2} + \frac{4 x ^{3}}{3} + 4 x^{2} + 16 x + 32 ln ∣ x - 2 ∣ + C

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What is the integral of (x^5)/(x-2) ?
The integral of (x^5)/(x-2) is (x^5)/5+(x^4)/2+(4x^3)/3+4x^2+16x+32ln|x-2|+C