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

The integral of (\sqrt[3]{x}-1/x)^2 is -1/x-6\sqrt[3]{x}+3/5 x^{5/3}+C

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

Solution steps

Expand

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

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

\int \frac{2}{x ^{\frac{2}{3}}} d x = 6 x^{\frac{1}{3}}

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

= - \frac{1}{x} - 6 x^{\frac{1}{3}} + \frac{3}{5} x^{\frac{5}{3}}

Apply radical rule:

Add a constant to the solution

\frac{d}{d x} (arctan (\frac{x}{4}))

f (x) = \frac{1}{x + 1}, (0, 1)

y = 6 e^{x}, y = 6 e^{- x}, x = 1

\int \frac{1}{y ^{2} - 9} d y

\int \frac{1}{9 - 64 x ^{2}} d x

What is the integral of (\sqrt[3]{x}-1/x)^2 ?
The integral of (\sqrt[3]{x}-1/x)^2 is -1/x-6\sqrt[3]{x}+3/5 x^{5/3}+C