What is the integral of 9/(sqrt(-x^2+2x)) ?

The integral of 9/(sqrt(-x^2+2x)) is 9arcsin(x-1)+C

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integral of 9/(sqrt(-x^2+2x))

\int \frac{9}{- x ^{2} + 2 x} d x

9 arcsin (x - 1) + C

Solution steps

\int \frac{9}{- x ^{2} + 2 x} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{- x ^{2} + 2 x} d x

Complete the square

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

= 9 \cdot \int \frac{1}{- ( x - 1 ) ^{2} + 1} d x

Apply u-substitution

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

= 9 \cdot \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

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

= 9 arcsin (u)

Substitute back

u = x - 1

= 9 arcsin (x - 1)

Add a constant to the solution

= 9 arcsin (x - 1) + C

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What is the integral of 9/(sqrt(-x^2+2x)) ?
The integral of 9/(sqrt(-x^2+2x)) is 9arcsin(x-1)+C