What is the integral of 6x^2-8x+3 ?

The integral of 6x^2-8x+3 is 2x^3-4x^2+3x+C

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integral of 6x^2-8x+3

\int 6 x^{2} - 8 x + 3 d x

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

Solution steps

\int 6 x^{2} - 8 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 6 x^{2} d x - \int 8 x d x + \int 3 d x

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

\int 8 x d x = 4 x^{2}

\int 3 d x = 3 x

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

Add a constant to the solution

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

x^{3} (3 - x)^{4}

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

\frac{d}{d x} (100 + x^{2})

y = x 6^{x}

x (x + y) (\frac{d y}{d x}) + y (3 x + y) = 0