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

The integral of (x+4)^2(x-4) is (x^4)/4+(4x^3)/3-8x^2-64x+C

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

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

\frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} - 8 x^{2} - 64 x + C

Solution steps

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

Expand

(x + 4)^{2} (x - 4) : x^{3} + 4 x^{2} - 16 x - 64

= \int x^{3} + 4 x^{2} - 16 x - 64 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 4 x^{2} d x - \int 16 x d x - \int 64 d x

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

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 16 x d x = 8 x^{2}

\int 64 d x = 64 x

= \frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} - 8 x^{2} - 64 x

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} - 8 x^{2} - 64 x + C

x^{2}

e^{t}

\frac{d}{d x} (\frac{250000 x ^{2}}{( 2 x + 1 ) ^{2}})

\frac{d ^{2}}{d x ^{2}} (x^{2} e^{- x^{2}})

\int 3 + 2 x y^{2} d x

What is the integral of (x+4)^2(x-4) ?
The integral of (x+4)^2(x-4) is (x^4)/4+(4x^3)/3-8x^2-64x+C