What is the integral of-3x^2+24x-41 ?

The integral of-3x^2+24x-41 is -x^3+12x^2-41x+C

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integral of-3x^2+24x-41

\int - 3 x^{2} + 24 x - 41 d x

- x^{3} + 12 x^{2} - 41 x + C

Solution steps

\int - 3 x^{2} + 24 x - 41 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 3 x^{2} d x + \int 24 x d x - \int 41 d x

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

\int 24 x d x = 12 x^{2}

\int 41 d x = 41 x

= - x^{3} + 12 x^{2} - 41 x

Add a constant to the solution

= - x^{3} + 12 x^{2} - 41 x + C

\int \frac{x}{1 + x} d x

\int 2 x x - 5 d x

\int 12 x + sin (x) d x

\int \frac{1}{x ( in x ) ^{2} + 9} d x

\int (1 - x) d x

What is the integral of-3x^2+24x-41 ?
The integral of-3x^2+24x-41 is -x^3+12x^2-41x+C