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

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

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

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

arcsin (x - 1) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Use the common integral:

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

= arcsin (u)

Substitute back

u = x - 1

= arcsin (x - 1)

Add a constant to the solution

= arcsin (x - 1) + C

\int (x^{6} + 1) d x

x \to 0 + lim (x^{2 x})

y^{^{'}} = \frac{2 x}{( y + x ^{2} y )}

y \to 0 lim (\frac{sin ( y )}{y})

\frac{1}{x + k e ^{x}}

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