What is the integral of (x+3)(x-3) ?

The integral of (x+3)(x-3) is (x^3)/3-9x+C

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integral of (x+3)(x-3)

\int (x + 3) (x - 3) d x

\frac{x ^{3}}{3} - 9 x + C

Solution steps

\int (x + 3) (x - 3) d x

Expand

(x + 3) (x - 3) : x^{2} - 9

= \int x^{2} - 9 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 9 d x

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

\int 9 d x = 9 x

= \frac{x ^{3}}{3} - 9 x

Add a constant to the solution

= \frac{x ^{3}}{3} - 9 x + C

y^{^{''}} - 2 y^{^{'}} = x + 2 e^{x}, y (0) = 17, y^{^{'}} (0) = \frac{55}{4}

\frac{1}{s ^{2} - 2 s + 26}

\frac{\partial}{\partial y} (ln (1 + x^{2} + e^{y} + z))

\frac{s ^{2} + 6 s + 2}{s ( s + 1 ) ^{2}}

y^{^{'}} = t y - 2 t + y - 2