What is the integral of (x^3-6x-9) ?

The integral of (x^3-6x-9) is (x^4)/4-3x^2-9x+C

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integral of (x^3-6x-9)

\int (x^{3} - 6 x - 9) d x

\frac{x ^{4}}{4} - 3 x^{2} - 9 x + C

Solution steps

\int x^{3} - 6 x - 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^{3} d x - \int 6 x d x - \int 9 d x

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

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

\int 9 d x = 9 x

= \frac{x ^{4}}{4} - 3 x^{2} - 9 x

Add a constant to the solution

= \frac{x ^{4}}{4} - 3 x^{2} - 9 x + C

f (x) = \frac{4 x}{( x - 9 ) ^{5}}

\frac{\partial}{\partial z} (x y e^{x z})

ln (4 x), 1

\int \frac{1}{( x ^{2} - 64 ) ^{2}} d x

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

What is the integral of (x^3-6x-9) ?
The integral of (x^3-6x-9) is (x^4)/4-3x^2-9x+C