What is the integral of (8x^3)/(3y^2) ?

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

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integral of (8x^3)/(3y^2)

\int \frac{8 x ^{3}}{3 y ^{2}}

\frac{2 x ^{4}}{3 y ^{2}} + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

: \frac{x ^{4}}{4}

= \frac{8}{3 y ^{2}} \cdot \frac{x ^{4}}{4}

Simplify

\frac{8}{3 y ^{2}} \cdot \frac{x ^{4}}{4} : \frac{2 x ^{4}}{3 y ^{2}}

= \frac{2 x ^{4}}{3 y ^{2}}

Add a constant to the solution

= \frac{2 x ^{4}}{3 y ^{2}} + C

\frac{\partial}{\partial x} (e^{y} + x)

\frac{d}{d x} ((6^{\frac{2}{3}} - x^{\frac{2}{3}})^{\frac{3}{2}})

\int \frac{1}{x ^{2} + x + 5} d x

\int e^{3 x - 2} d x

\frac{d y}{d x} - 9 y^{2} e^{- 1 x} = 0, y (0) = 5

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