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

The integral of (x-4)(x+4) is (x^3)/3-16x+C

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

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

\frac{x ^{3}}{3} - 16 x + C

Solution steps

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

Expand

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

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

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

\int 16 d x = 16 x

= \frac{x ^{3}}{3} - 16 x

Add a constant to the solution

= \frac{x ^{3}}{3} - 16 x + C

\int \frac{( arctan ( x ) ) ^{6}}{x ^{2} + 1} d x

\int \frac{1}{( 81 + x ^{2} ) ^{2}} d x

y^{^{''}} - 6 y^{^{'}} + 9 y = 5 t^{- 1} e^{3 t}

\int \frac{x ^{2}}{5 - x ^{3}} d x

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

What is the integral of (x-4)(x+4) ?
The integral of (x-4)(x+4) is (x^3)/3-16x+C