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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{u + 1}{u ^{2} - u ^{2} + 2} : \frac{1}{u - u ^{2} + 2} + \frac{1}{u ^{2} - u ^{2} + 2}

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

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

\int \frac{1}{u - u ^{2} + 2} d u = - \frac{1}{2 2} (ln \frac{- u ^{2} + 2}{2} + 1 - ln \frac{- u ^{2} + 2}{2} - 1)

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

= - \frac{1}{2 2} (ln \frac{- u ^{2} + 2}{2} + 1 - ln \frac{- u ^{2} + 2}{2} - 1) - \frac{2 - u ^{2}}{2 u}

Substitute back

u = x - 1

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

Simplify

- \frac{1}{2 2} ln \frac{- ( x - 1 ) ^{2} + 2}{2} + 1 - ln \frac{- ( x - 1 ) ^{2} + 2}{2} - 1 - \frac{2 - ( x - 1 ) ^{2}}{2 ( x - 1 )} : - \frac{1}{2 2} (ln \frac{- x ^{2} + 2 x + 1}{2} + 1 - ln \frac{- x ^{2} + 2 x + 1}{2} - 1) - \frac{- x ^{2} + 2 x + 1}{2 ( x - 1 )}

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

Add a constant to the solution

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

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