What is the integral of (2x-4)(x+2) ?

The integral of (2x-4)(x+2) is (2x^3)/3-8x+C

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integral of (2x-4)(x+2)

\int (2 x - 4) (x + 2) d x

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

Solution steps

\int (2 x - 4) (x + 2) d x

Expand

(2 x - 4) (x + 2) : 2 x^{2} - 8

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

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

\int 8 d x = 8 x

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

Add a constant to the solution

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

x - 4, x = 8

\frac{\partial}{\partial x} (e^{x} + z^{2})

2 (y - 4 x^{2}) d x + x d y = 0

\int \frac{1}{cos ^{2} ( x ) + 3 sin ^{2} ( x )} d x

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

What is the integral of (2x-4)(x+2) ?
The integral of (2x-4)(x+2) is (2x^3)/3-8x+C