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

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

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

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

ln ∣ x ∣ + ln 1 - x + 1 - ln 1 - x - 1 - 2 1 - x + C

Solution steps

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

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

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= ln ∣ x ∣ - (- ln 1 - x + 1 + ln 1 - x - 1 + 2 1 - x)

Simplify

= ln ∣ x ∣ + ln 1 - x + 1 - ln 1 - x - 1 - 2 1 - x

Add a constant to the solution

= ln ∣ x ∣ + ln 1 - x + 1 - ln 1 - x - 1 - 2 1 - x + C

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