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

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

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

\frac{x ^{3}}{3} - \frac{3}{2 x ^{\frac{2}{3}}} + C

Solution steps

Expand

= \int x^{2} + \frac{1}{x ^{\frac{5}{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^{2} d x + \int \frac{1}{x ^{\frac{5}{3}}} d x

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

\int \frac{1}{x ^{\frac{5}{3}}} d x = - \frac{3}{2 x ^{\frac{2}{3}}}

= \frac{x ^{3}}{3} - \frac{3}{2 x ^{\frac{2}{3}}}

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{3}{2 x ^{\frac{2}{3}}} + C

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