What is the integral of 1/(sqrt((2x-x^2)^3)) ?

The integral of 1/(sqrt((2x-x^2)^3)) is (x-1)/(sqrt(-x^2+2x))+C

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integral of 1/(sqrt((2x-x^2)^3))

\int \frac{1}{( 2 x - x ^{2} ) ^{3}} d x

\frac{x - 1}{- x ^{2} + 2 x} + C

Solution steps

\int \frac{1}{( 2 x - x ^{2} ) ^{3}} d x

Complete the square

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

= \int \frac{1}{( - ( x - 1 ) ^{2} + 1 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{( - u ^{2} + 1 ) ^{\frac{3}{2}}} d u

= \int \frac{1}{( - u ^{2} + 1 ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int sec^{2} (v) d v

= \int sec^{2} (v) d v

Use the common integral:

\int sec^{2} (v) d v = tan (v)

= tan (v)

Substitute back

= tan (arcsin (x - 1))

Simplify

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

= \frac{x - 1}{- x ^{2} + 2 x}

Add a constant to the solution

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

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What is the integral of 1/(sqrt((2x-x^2)^3)) ?
The integral of 1/(sqrt((2x-x^2)^3)) is (x-1)/(sqrt(-x^2+2x))+C