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

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

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integral of (4x)/3+y^2

\int \frac{4 x}{3} + y^{2} d y

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

Solution steps

\int \frac{4 x}{3} + y^{2} d y

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}{3} d y + \int y^{2} d y

\int \frac{4 x}{3} d y = \frac{4 x}{3} y

\int y^{2} d y = \frac{y ^{3}}{3}

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

Add a constant to the solution

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

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