What is the integral of 1/((x^2-2x+1)sqrt(x^2-2x)) ?

The integral of 1/((x^2-2x+1)sqrt(x^2-2x)) is (sqrt(x^2-2x))/(x-1)+C

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integral of 1/((x^2-2x+1)sqrt(x^2-2x))

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

\frac{x ^{2} - 2 x}{x - 1} + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{1}{u ^{2} u ^{2} - 1} d u

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

Apply Trigonometric Substitution

: \int cos (v) d v

= \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= sin (v)

Substitute back

= sin (arcsec (x - 1))

Simplify

sin (arcsec (x - 1)) : \frac{x ^{2} - 2 x}{x - 1}

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

Add a constant to the solution

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

\frac{\partial}{\partial z} (2 x^{2} y)

\frac{d}{d x} (2 x^{5} + 6 x^{8})

\int_{0}^{1} (7 x + x) d x

f (x) = x^{5} + 5^{x}

f (x) = - 4 6 x^{2} - 5 x + 10

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