What is the integral of \sqrt[3]{x}-1/(x^2) ?

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

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integral of \sqrt[3]{x}-1/(x^2)

\frac{3}{4} x^{\frac{4}{3}} + \frac{1}{x} + 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}{x ^{2}} d x = - \frac{1}{x}

= \frac{3}{4} x^{\frac{4}{3}} - (- \frac{1}{x})

Simplify

= \frac{3}{4} x^{\frac{4}{3}} + \frac{1}{x}

Add a constant to the solution

= \frac{3}{4} x^{\frac{4}{3}} + \frac{1}{x} + C

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