What is the integral of (x^2-2x-3) ?

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

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integral of (x^2-2x-3)

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

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

Solution steps

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

(x^{2} + 1) y^{^{'}} + y^{2} + 1 = 0, y (4) = e

\int \frac{1}{( sin ( x ) ) ^{2} ( cos ( x ) ) ^{4}} d x

\frac{d ^{4}}{d x ^{4}} (ln (x))

15 x^{\frac{1}{4}}

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