What is the integral of (x+2)(x-5) ?

The integral of (x+2)(x-5) is (x^3)/3-(3x^2)/2-10x+C

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integral of (x+2)(x-5)

\int (x + 2) (x - 5) d x

\frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} - 10 x + C

Solution steps

\int (x + 2) (x - 5) d x

Expand

(x + 2) (x - 5) : x^{2} - 3 x - 10

= \int x^{2} - 3 x - 10 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 3 x d x - \int 10 d x

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

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

\int 10 d x = 10 x

= \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} - 10 x

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} - 10 x + C

\frac{d}{d x} (e^{\frac{x ^{4}}{10}})

\frac{d y}{d x} = x^{2} + y - 1

y^{^{''}} + 9 y = 2 sin (2 x), y (0) = 0, y^{^{'}} (0) = - 1

\frac{4 x}{( x ^{2} + 1 )}

\int_{0}^{3} (1 + 6 x^{2} - 10 x^{4}) d x

What is the integral of (x+2)(x-5) ?
The integral of (x+2)(x-5) is (x^3)/3-(3x^2)/2-10x+C