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

The integral of (x^2-3)(x+2) is (x^4)/3-3x^2+(2x^3)/3-6x-(x^4}{12}+\frac{3x^2)/2+C

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

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

\frac{x ^{4}}{3} - 3 x^{2} + \frac{2 x ^{3}}{3} - 6 x - \frac{x ^{4}}{12} + \frac{3 x ^{2}}{2} + C

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

(x + 2) (\frac{x ^{3}}{3} - 3 x) - (\frac{x ^{4}}{12} - \frac{3 x ^{2}}{2}) : \frac{x ^{4}}{3} - 3 x^{2} + \frac{2 x ^{3}}{3} - 6 x - \frac{x ^{4}}{12} + \frac{3 x ^{2}}{2}

= \frac{x ^{4}}{3} - 3 x^{2} + \frac{2 x ^{3}}{3} - 6 x - \frac{x ^{4}}{12} + \frac{3 x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{4}}{3} - 3 x^{2} + \frac{2 x ^{3}}{3} - 6 x - \frac{x ^{4}}{12} + \frac{3 x ^{2}}{2} + C

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What is the integral of (x^2-3)(x+2) ?
The integral of (x^2-3)(x+2) is (x^4)/3-3x^2+(2x^3)/3-6x-(x^4}{12}+\frac{3x^2)/2+C