What is the integral of (\sqrt[3]{u}+1/(sqrt(u))) ?

The integral of (\sqrt[3]{u}+1/(sqrt(u))) is 3/4 u^{4/3}+2sqrt(u)+C

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integral of (\sqrt[3]{u}+1/(sqrt(u)))

\frac{3}{4} u^{\frac{4}{3}} + 2 u + C

Solution steps

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{1}{u} d u = 2 u

= \frac{3}{4} u^{\frac{4}{3}} + 2 u

Add a constant to the solution

= \frac{3}{4} u^{\frac{4}{3}} + 2 u + C

- x^{2} - 1, x^{2} + 4, [- 2, 3]

y^{^{'}} = x (1 + y)

\frac{\partial}{\partial x} (t x)

\int - \frac{1}{2} x d x

\int \frac{x ^{3}}{6} d x

What is the integral of (\sqrt[3]{u}+1/(sqrt(u))) ?
The integral of (\sqrt[3]{u}+1/(sqrt(u))) is 3/4 u^{4/3}+2sqrt(u)+C