What is the integral of 20x^2+3x-9 ?

The integral of 20x^2+3x-9 is (20x^3)/3+(3x^2)/2-9x+C

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integral of 20x^2+3x-9

\int 20 x^{2} + 3 x - 9 d x

\frac{20 x ^{3}}{3} + \frac{3 x ^{2}}{2} - 9 x + C

Solution steps

\int 20 x^{2} + 3 x - 9 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 20 x^{2} d x + \int 3 x d x - \int 9 d x

\int 20 x^{2} d x = \frac{20 x ^{3}}{3}

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

\int 9 d x = 9 x

= \frac{20 x ^{3}}{3} + \frac{3 x ^{2}}{2} - 9 x

Add a constant to the solution

= \frac{20 x ^{3}}{3} + \frac{3 x ^{2}}{2} - 9 x + C

\int (6 x^{3} + 3 x^{2}) d x

ln (sec (x))

x \to 0 lim (x^{2} cos (\frac{6}{x}))

\frac{d}{d x} (\frac{3}{2})

\frac{d}{d x} ((1 + 4 x^{3}) (1 + 2 x^{2}))

What is the integral of 20x^2+3x-9 ?
The integral of 20x^2+3x-9 is (20x^3)/3+(3x^2)/2-9x+C