What is the integral of (9x^2-36x)/4+18 ?

The integral of (9x^2-36x)/4+18 is 1/4 (3x^3-18x^2)+18x+C

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integral of (9x^2-36x)/4+18

\int \frac{9 x ^{2} - 36 x}{4} + 18 d x

\frac{1}{4} (3 x^{3} - 18 x^{2}) + 18 x + C

Solution steps

\int \frac{9 x ^{2} - 36 x}{4} + 18 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 \frac{9 x ^{2} - 36 x}{4} d x + \int 18 d x

\int \frac{9 x ^{2} - 36 x}{4} d x = \frac{1}{4} (3 x^{3} - 18 x^{2})

\int 18 d x = 18 x

= \frac{1}{4} (3 x^{3} - 18 x^{2}) + 18 x

Add a constant to the solution

= \frac{1}{4} (3 x^{3} - 18 x^{2}) + 18 x + C

\frac{d y}{d x} = cos (x + y)

\int \frac{x ^{4} + 1}{x ^{4} - 1} d x

\int \frac{2 x - 1}{( x - 1 ) ( x - 2 ) ( 2 x - 3 )} d x

\frac{d}{d x} (e^{- x} cos (y))

\frac{d}{d x} ((1 + x^{3})^{5})

What is the integral of (9x^2-36x)/4+18 ?
The integral of (9x^2-36x)/4+18 is 1/4 (3x^3-18x^2)+18x+C