What is the integral of (x^2+4x-5) ?

The integral of (x^2+4x-5) is (x^3)/3+2x^2-5x+C

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integral of (x^2+4x-5)

\int (x^{2} + 4 x - 5) d x

\frac{x ^{3}}{3} + 2 x^{2} - 5 x + C

Solution steps

\int x^{2} + 4 x - 5 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 5 d x

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

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

\int 5 d x = 5 x

= \frac{x ^{3}}{3} + 2 x^{2} - 5 x

Add a constant to the solution

= \frac{x ^{3}}{3} + 2 x^{2} - 5 x + C

What is the integral of (x^2+4x-5) ?
The integral of (x^2+4x-5) is (x^3)/3+2x^2-5x+C