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

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

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{y}{y ^{2} + x ^{2}})

\int_{- 1}^{1} x cos (nπ x) d x

\int \frac{1}{1 - v ^{2}} d v

\int (x - 2)^{2} d x

\frac{d}{d x} (\frac{1 - ln ( x )}{1 + ln ( x )})

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