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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = 1 + x

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

Simplify

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

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

Add a constant to the solution

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

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