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

\int (3 x - \frac{1}{x})^{2} d x

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

Solution steps

\int (3 x - \frac{1}{x})^{2} d x

Expand

(3 x - \frac{1}{x})^{2} : \frac{1}{x ^{2}} - \frac{2}{x ^{\frac{2}{3}}} + x^{\frac{2}{3}}

= \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:

a^{\frac{1}{n}} = n a

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

Add a constant to the solution

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

t an g e n t f (x) = \frac{1}{x + 1}, (0, 1)

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

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

d er i v a t i v e y (θ) = π^{4} cos (θ)

\frac{d}{d t} (9 + t)

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