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

\int_{0}^{1} \frac{1 + 12 t}{1 + 3 t} d t

(- 6 sin (6 x))^{^{'}}

\frac{d}{d x} (\frac{1}{8 x + 3})

\int (- 5 e^{x} + \frac{4}{x}) d x

ln (\frac{a - b x}{a - c x})

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