What is the integral of (x+1)(x-5) ?

The integral of (x+1)(x-5) is (x^3)/3-2x^2-5x+C

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integral of (x+1)(x-5)

\int (x + 1) (x - 5) d x

\frac{x ^{3}}{3} - 2 x^{2} - 5 x + C

Solution steps

\int (x + 1) (x - 5) d x

Expand

(x + 1) (x - 5) : x^{2} - 4 x - 5

= \int x^{2} - 4 x - 5 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 4 x d x - \int 5 d x

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

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

\int 5 d x = 5 x

= \frac{x ^{3}}{3} - 2 x^{2} - 5 x

Add a constant to the solution

= \frac{x ^{3}}{3} - 2 x^{2} - 5 x + C

\int (\frac{8 x ^{3}}{3} - 3 x^{2} + 5) d x

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What is the integral of (x+1)(x-5) ?
The integral of (x+1)(x-5) is (x^3)/3-2x^2-5x+C