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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{r}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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