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

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

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

- \frac{5}{2} (1 - x)^{\frac{2}{5}} + C

Solution steps

Apply u-substitution

: \int - \frac{1}{u ^{\frac{3}{5}}} d u

= \int - \frac{1}{u ^{\frac{3}{5}}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{u ^{\frac{3}{5}}} d u

Apply the Power Rule

: \frac{5}{2} u^{\frac{2}{5}}

= - \frac{5}{2} u^{\frac{2}{5}}

Substitute back

u = 1 - x

= - \frac{5}{2} (1 - x)^{\frac{2}{5}}

Add a constant to the solution

= - \frac{5}{2} (1 - x)^{\frac{2}{5}} + C

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