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

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

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

\int (x + 1) (x - 2) d x

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

Solution steps

\int (x + 1) (x - 2) d x

Expand

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

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

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

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

\int 2 d x = 2 x

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

Add a constant to the solution

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

f (x) = \frac{1 - 5 x}{4 + x}

\frac{\partial}{\partial x} (π^{7})

\frac{d}{d x} (2 sin^{2} (2 x))

f (x) = \frac{3}{x - 6}

x \to - 2 - lim (\frac{x ^{2}}{x + 2})

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