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

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

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integral of 2/((y-1)sqrt(y^2-2y))

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

2 arctan (y^{2} - 2 y) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{v ^{2} + 1} d v

= 2 \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 arctan (v)

Substitute back

= 2 arctan ((y - 1)^{2} - 1)

Expand

(y - 1)^{2} - 1 : y^{2} - 2 y

= 2 arctan (y^{2} - 2 y)

Add a constant to the solution

= 2 arctan (y^{2} - 2 y) + C

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