What is the integral of 4x^3-6x^2+3 ?

The integral of 4x^3-6x^2+3 is x^4-2x^3+3x+C

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integral of 4x^3-6x^2+3

\int 4 x^{3} - 6 x^{2} + 3 d x

x^{4} - 2 x^{3} + 3 x + C

Solution steps

\int 4 x^{3} - 6 x^{2} + 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 4 x^{3} d x - \int 6 x^{2} d x + \int 3 d x

\int 4 x^{3} d x = x^{4}

\int 6 x^{2} d x = 2 x^{3}

\int 3 d x = 3 x

= x^{4} - 2 x^{3} + 3 x

Add a constant to the solution

= x^{4} - 2 x^{3} + 3 x + C

\int 19 tan^{5} (x) sec^{4} (x) d x

f (x) = 2 e^{- 3 x + 9}

(- 2, - 1), (8, - 3)

\int \frac{sec ( x )}{tan ( x ) + cot ( x )} d x

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

What is the integral of 4x^3-6x^2+3 ?
The integral of 4x^3-6x^2+3 is x^4-2x^3+3x+C