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

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

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

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

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

Solution steps

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

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

x \to \frac{1}{2} lim (x sec (π) x)

\int 8 e^{8 x} d x

81 y^{^{''}} + 90 y^{^{'}} + 25 y = 0

\frac{d}{d x} (5 x^{2} + 10)

(5 x^{2} - 10 x + 10) e^{x}

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