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

The integral of (2/(\sqrt[5]{x)}-1)^2 is 20/3 x^{3/5}-5x^{4/5}+x+C

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integral of (2/(\sqrt[5]{x)}-1)^2

\frac{20}{3} x^{\frac{3}{5}} - 5 x^{\frac{4}{5}} + x + C

Solution steps

Expand

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{4}{x ^{\frac{2}{5}}} d x = \frac{20}{3} x^{\frac{3}{5}}

\int 1 d x = x

= \frac{20}{3} x^{\frac{3}{5}} - 5 x^{\frac{4}{5}} + x

Add a constant to the solution

= \frac{20}{3} x^{\frac{3}{5}} - 5 x^{\frac{4}{5}} + x + C

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What is the integral of (2/(\sqrt[5]{x)}-1)^2 ?
The integral of (2/(\sqrt[5]{x)}-1)^2 is 20/3 x^{3/5}-5x^{4/5}+x+C