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

The integral of (x^2+4\sqrt[3]{x})/x is (x^2)/2+12\sqrt[3]{x}+C

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integral of (x^2+4\sqrt[3]{x})/x

Solution steps

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int x d x = \frac{x ^{2}}{2}

\int \frac{4}{x ^{\frac{2}{3}}} d x = 12 x^{\frac{1}{3}}

= \frac{x ^{2}}{2} + 12 x^{\frac{1}{3}}

Apply radical rule:

Add a constant to the solution

What is the integral of (x^2+4\sqrt[3]{x})/x ?
The integral of (x^2+4\sqrt[3]{x})/x is (x^2)/2+12\sqrt[3]{x}+C