What is the integral of ((2x^2)/7-8/3 x^4) ?

The integral of ((2x^2)/7-8/3 x^4) is (2x^3)/(21)-(8x^5)/(15)+C

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integral of ((2x^2)/7-8/3 x^4)

\int (\frac{2 x ^{2}}{7} - \frac{8}{3} x^{4}) d x

\frac{2 x ^{3}}{21} - \frac{8 x ^{5}}{15} + C

Solution steps

\int \frac{2 x ^{2}}{7} - \frac{8}{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 \frac{2 x ^{2}}{7} d x - \int \frac{8}{3} x^{4} d x

\int \frac{2 x ^{2}}{7} d x = \frac{2 x ^{3}}{21}

\int \frac{8}{3} x^{4} d x = \frac{8 x ^{5}}{15}

= \frac{2 x ^{3}}{21} - \frac{8 x ^{5}}{15}

Add a constant to the solution

= \frac{2 x ^{3}}{21} - \frac{8 x ^{5}}{15} + C

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What is the integral of ((2x^2)/7-8/3 x^4) ?
The integral of ((2x^2)/7-8/3 x^4) is (2x^3)/(21)-(8x^5)/(15)+C