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

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

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integral of ((x^2+1-2x)^{1/5})/(1-x)

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

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

Solution steps

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

Complete the square

x^{2} + 1 - 2 x : (x - 1)^{2}

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

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 = x - 1

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

Add a constant to the solution

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

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