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

y^{^{'}} = \frac{( 5 x ^{2} + y ^{2} )}{x y}

\int_{0}^{\frac{1}{2} - x} 384 x y d y

\int x (4 x^{2} + 7)^{9} d x

\frac{d y}{d x} = \frac{y}{1 + x ^{2}}

f^{^{''}} (x) = \frac{3}{4} x^{4} + 2 x^{3} - 12 x^{2} + 29

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