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

The integral of-4x^4+8x^3 is -(4x^5)/5+2x^4+C

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integral of-4x^4+8x^3

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

- \frac{4 x ^{5}}{5} + 2 x^{4} + C

Solution steps

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

\int 4 x^{4} d x = \frac{4 x ^{5}}{5}

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

= - \frac{4 x ^{5}}{5} + 2 x^{4}

Add a constant to the solution

= - \frac{4 x ^{5}}{5} + 2 x^{4} + C

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What is the integral of-4x^4+8x^3 ?
The integral of-4x^4+8x^3 is -(4x^5)/5+2x^4+C