What is the integral of (x^2-4)(x-2) ?

The integral of (x^2-4)(x-2) is (x^4)/3-(x^4}{12}-\frac{2x^3)/3-2x^2+8x+C

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integral of (x^2-4)(x-2)

\int (x^{2} - 4) (x - 2) d x

\frac{x ^{4}}{3} - \frac{x ^{4}}{12} - \frac{2 x ^{3}}{3} - 2 x^{2} + 8 x + C

Solution steps

\int (x^{2} - 4) (x - 2) d x

Apply Integration By Parts

: (x - 2) (\frac{x ^{3}}{3} - 4 x) - \int \frac{x ^{3}}{3} - 4 x d x

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

\int \frac{x ^{3}}{3} - 4 x d x = \frac{x ^{4}}{12} - 2 x^{2}

= (x - 2) (\frac{x ^{3}}{3} - 4 x) - (\frac{x ^{4}}{12} - 2 x^{2})

Simplify

(x - 2) (\frac{x ^{3}}{3} - 4 x) - (\frac{x ^{4}}{12} - 2 x^{2}) : \frac{x ^{4}}{3} - \frac{x ^{4}}{12} - \frac{2 x ^{3}}{3} - 2 x^{2} + 8 x

= \frac{x ^{4}}{3} - \frac{x ^{4}}{12} - \frac{2 x ^{3}}{3} - 2 x^{2} + 8 x

Add a constant to the solution

= \frac{x ^{4}}{3} - \frac{x ^{4}}{12} - \frac{2 x ^{3}}{3} - 2 x^{2} + 8 x + C

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What is the integral of (x^2-4)(x-2) ?
The integral of (x^2-4)(x-2) is (x^4)/3-(x^4}{12}-\frac{2x^3)/3-2x^2+8x+C