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

The integral of 1/((x+1)sqrt(1-x)) is -sqrt(2)(1/2 ln|sqrt((-x+1)/2)+1|-1/2 ln|sqrt((-x+1)/2)-1|)+C

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integral of 1/((x+1)sqrt(1-x))

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

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

Solution steps

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

Apply u-substitution

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

Apply u-substitution

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= - 2 \cdot \frac{1}{2} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2})

Substitute back

= - 2 \cdot \frac{1}{2} \frac{ln \frac{- ( x + 1 ) + 2}{2} + 1}{2} - \frac{ln \frac{- ( x + 1 ) + 2}{2} - 1}{2}

Simplify

- 2 \cdot \frac{1}{2} \frac{ln \frac{- ( x + 1 ) + 2}{2} + 1}{2} - \frac{ln \frac{- ( x + 1 ) + 2}{2} - 1}{2} : - 2 (\frac{1}{2} ln \frac{- x + 1}{2} + 1 - \frac{1}{2} ln \frac{- x + 1}{2} - 1)

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

Add a constant to the solution

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

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