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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{1}{sin ( v ) + 1} d v

= \int \frac{1}{sin ( v ) + 1} d v

Apply u-substitution

: \int \frac{2}{2 w + 1 + w ^{2}} d w

= \int \frac{2}{2 w + 1 + w ^{2}} d w

Take the constant out:

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

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

Complete the square

2 w + 1 + w^{2} : (w + 1)^{2}

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

Apply u-substitution

: \int \frac{1}{t ^{2}} d t

= 2 \cdot \int \frac{1}{t ^{2}} d t

Apply the Power Rule

: - \frac{1}{t}

= 2 (- \frac{1}{t})

Substitute back

= 2 - \frac{1}{tan ( \frac{a r c s i n ( x - 1 )}{2} ) + 1}

Simplify

2 - \frac{1}{tan ( \frac{a r c s i n ( x - 1 )}{2} ) + 1} : - \frac{2}{tan ( \frac{1}{2} arcsin ( x - 1 ) ) + 1}

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

Add a constant to the solution

= - \frac{2}{tan ( \frac{1}{2} arcsin ( x - 1 ) ) + 1} + C

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