What is the integral of 9x^2+4x-3 ?

The integral of 9x^2+4x-3 is 3x^3+2x^2-3x+C

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integral of 9x^2+4x-3

\int 9 x^{2} + 4 x - 3 d x

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

Solution steps

\int 9 x^{2} + 4 x - 3 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 9 x^{2} d x + \int 4 x d x - \int 3 d x

\int 9 x^{2} d x = 3 x^{3}

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

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

x \to 4 lim (2 x^{2} + 7)

\frac{d h}{d t} = \frac{( 7 - e ^{- \frac{t}{5}} )}{6} - \frac{2}{9} h

2 x y^{3} = x y + 6

\int \frac{1}{4 x} + 9 d x