What is the integral of (8/3+4u^2) ?

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

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integral of (8/3+4u^2)

\int (\frac{8}{3} + 4 u^{2}) d u

\frac{8}{3} u + \frac{4 u ^{3}}{3} + C

Solution steps

\int \frac{8}{3} + 4 u^{2} d u

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{8}{3} d u + \int 4 u^{2} d u

\int \frac{8}{3} d u = \frac{8}{3} u

\int 4 u^{2} d u = \frac{4 u ^{3}}{3}

= \frac{8}{3} u + \frac{4 u ^{3}}{3}

Add a constant to the solution

= \frac{8}{3} u + \frac{4 u ^{3}}{3} + C

\frac{d ^{2}}{d x ^{2}} (1 + x)

\frac{\partial}{\partial y} (cos (8 x) sin (2 y))

x^{4} - 3 x^{2} + 1

n = 0 \sum \infty \frac{1}{n ^{3} + 1}

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