What is the integral of 3x^2+4x+8 ?

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

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integral of 3x^2+4x+8

\int 3 x^{2} + 4 x + 8 d x

x^{3} + 2 x^{2} + 8 x + C

Solution steps

\int 3 x^{2} + 4 x + 8 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 3 x^{2} d x + \int 4 x d x + \int 8 d x

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

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

\int 8 d x = 8 x

= x^{3} + 2 x^{2} + 8 x

Add a constant to the solution

= x^{3} + 2 x^{2} + 8 x + C

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