What is the integral of \sqrt[3]{w}+10\sqrt[5]{w^3} ?

The integral of \sqrt[3]{w}+10\sqrt[5]{w^3} is 3/4 w^{4/3}+25/4 w^{8/5}+C

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integral of \sqrt[3]{w}+10\sqrt[5]{w^3}

\frac{3}{4} w^{\frac{4}{3}} + \frac{25}{4} w^{\frac{8}{5}} + 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

= \frac{3}{4} w^{\frac{4}{3}} + \frac{25}{4} w^{\frac{8}{5}}

Add a constant to the solution

= \frac{3}{4} w^{\frac{4}{3}} + \frac{25}{4} w^{\frac{8}{5}} + C

x \to - 1 - lim (- 4 x - 4 - 8)

y^{^{'}} + y = \frac{1 - x}{2 x ^{2}}

y^{6} + x^{3} = y^{2} + 12 x, (0, 1)

x \to - 1 lim (\frac{x}{0})

f (x) = 5 \cdot 3^{x}, at x = 2

What is the integral of \sqrt[3]{w}+10\sqrt[5]{w^3} ?
The integral of \sqrt[3]{w}+10\sqrt[5]{w^3} is 3/4 w^{4/3}+25/4 w^{8/5}+C