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

The integral of (x+4)^2(x-2) is (x^4)/4+2x^3-32x+C

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

\int (x + 4)^{2} (x - 2) d x

\frac{x ^{4}}{4} + 2 x^{3} - 32 x + C

Solution steps

\int (x + 4)^{2} (x - 2) d x

Expand

(x + 4)^{2} (x - 2) : x^{3} + 6 x^{2} - 32

= \int x^{3} + 6 x^{2} - 32 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^{3} d x + \int 6 x^{2} d x - \int 32 d x

\int x^{3} d x = \frac{x ^{4}}{4}

\int 6 x^{2} d x = 2 x^{3}

\int 32 d x = 32 x

= \frac{x ^{4}}{4} + 2 x^{3} - 32 x

Add a constant to the solution

= \frac{x ^{4}}{4} + 2 x^{3} - 32 x + C

f (x) = arcsec (x) + arccsc (x)

\frac{d}{d x} (\frac{x ^{3}}{30} + \frac{5}{2 x})

\int cot^{3} (7 x) d x

R (a) = (5 a + 1)^{2}

\frac{d}{d x} (y \cdot e^{x})

What is the integral of (x+4)^2(x-2) ?
The integral of (x+4)^2(x-2) is (x^4)/4+2x^3-32x+C