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

The integral of (((2x)/7)^2-8/3 x^4) is (4x^3)/(147)-(8x^5)/(15)+C

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

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

\frac{4 x ^{3}}{147} - \frac{8 x ^{5}}{15} + C

Solution steps

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

Simplify

(\frac{2 x}{7})^{2} - \frac{8}{3} x^{4} : \frac{4 x ^{2}}{49} - \frac{8}{3} x^{4}

= \int \frac{4 x ^{2}}{49} - \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{4 x ^{2}}{49} d x - \int \frac{8}{3} x^{4} d x

\int \frac{4 x ^{2}}{49} d x = \frac{4 x ^{3}}{147}

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

= \frac{4 x ^{3}}{147} - \frac{8 x ^{5}}{15}

Add a constant to the solution

= \frac{4 x ^{3}}{147} - \frac{8 x ^{5}}{15} + C

What is the integral of (((2x)/7)^2-8/3 x^4) ?
The integral of (((2x)/7)^2-8/3 x^4) is (4x^3)/(147)-(8x^5)/(15)+C