What is the integral of-x^2+12x+89 ?

The integral of-x^2+12x+89 is -(x^3)/3+6x^2+89x+C

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integral of-x^2+12x+89

\int - x^{2} + 12 x + 89 d x

- \frac{x ^{3}}{3} + 6 x^{2} + 89 x + C

Solution steps

\int - x^{2} + 12 x + 89 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 12 x d x + \int 89 d x

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

\int 12 x d x = 6 x^{2}

\int 89 d x = 89 x

= - \frac{x ^{3}}{3} + 6 x^{2} + 89 x

Add a constant to the solution

= - \frac{x ^{3}}{3} + 6 x^{2} + 89 x + C

f (x) = 2 x

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

\int \frac{2 x - 6}{4 x ^{2} - 24 x + 32} d x

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

\int (2 sin (x) + 3 sinh (x)) d x

What is the integral of-x^2+12x+89 ?
The integral of-x^2+12x+89 is -(x^3)/3+6x^2+89x+C