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

x \to - \infty lim (xx)

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

\frac{s + 7}{( s + 2 ) ( s - 5 )}

y = \frac{e ^{x}}{ln ( x )}

f (x) = (8 x^{3} + 2 x^{2})^{4}

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