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

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

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

\int \frac{x - 1}{2 x - x ^{2}} d x

- - x^{2} + 2 x + C

Solution steps

\int \frac{x - 1}{2 x - x ^{2}} d x

Expand

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

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

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{x}{2 x - x ^{2}} d x - \int \frac{1}{2 x - x ^{2}} d x

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

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

= - - x^{2} + 2 x + arcsin (x - 1) - arcsin (x - 1)

Simplify

= - - x^{2} + 2 x

Add a constant to the solution

= - - x^{2} + 2 x + C

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