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

The integral of (x-1)^2(x+2) is (x^4)/3-x^3+(2x^3)/3-x^2+2x-(x^4}{12}+\frac{x^3)/3-(x^2)/2+C

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integral of (x-1)^2(x+2)

\int (x - 1)^{2} (x + 2) d x

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

Solution steps

\int (x - 1)^{2} (x + 2) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (x-1)^2(x+2) ?
The integral of (x-1)^2(x+2) is (x^4)/3-x^3+(2x^3)/3-x^2+2x-(x^4}{12}+\frac{x^3)/3-(x^2)/2+C