What is the integral of (x^2-4x-15) ?

The integral of (x^2-4x-15) is (x^3)/3-2x^2-15x+C

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integral of (x^2-4x-15)

\int (x^{2} - 4 x - 15) d x

\frac{x ^{3}}{3} - 2 x^{2} - 15 x + C

Solution steps

\int x^{2} - 4 x - 15 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 4 x d x - \int 15 d x

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

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

\int 15 d x = 15 x

= \frac{x ^{3}}{3} - 2 x^{2} - 15 x

Add a constant to the solution

= \frac{x ^{3}}{3} - 2 x^{2} - 15 x + C

\int x^{2} cos (4 x) d x

\int x + y d x

f (x) = sin (8 e^{x})

\int (3 x^{2} y) d x

sin^{2} (x), sin^{3} (x), [0, 2 π]

What is the integral of (x^2-4x-15) ?
The integral of (x^2-4x-15) is (x^3)/3-2x^2-15x+C