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

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

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

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

2 arcsin (x - 1) + C

Solution steps

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

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

Use the common integral:

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

= 2 arcsin (u)

Substitute back

u = x - 1

= 2 arcsin (x - 1)

Add a constant to the solution

= 2 arcsin (x - 1) + C

g (x) = 4 - x

\frac{A}{y ^{13}} + B e^{y}

x = - 2, x = 1, y = 2 x^{2} + 11, y = 0

\int 16 sin^{2} (x) cos^{2} (x) d x

\int_{0}^{5} (7 x - x^{2}) - (2 x) d x

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