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

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

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integral of 4x^3+8x^2

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

x^{4} + \frac{8 x ^{3}}{3} + C

Solution steps

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

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

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

= x^{4} + \frac{8 x ^{3}}{3}

Add a constant to the solution

= x^{4} + \frac{8 x ^{3}}{3} + C

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