What is the integral of (x^2+2x-8) ?

The integral of (x^2+2x-8) is (x^3)/3+x^2-8x+C

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integral of (x^2+2x-8)

\int (x^{2} + 2 x - 8) d x

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

Solution steps

\int x^{2} + 2 x - 8 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 8 d x

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

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

\int 8 d x = 8 x

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

Add a constant to the solution

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

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What is the integral of (x^2+2x-8) ?
The integral of (x^2+2x-8) is (x^3)/3+x^2-8x+C