What is the integral of ln|1-x| ?

The integral of ln|1-x| is xln|1-x|+1-x-ln|1-x|+C

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integral of ln|1-x|

\int ln ∣ 1 - x ∣ d x

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

Solution steps

\int ln ∣ 1 - x ∣ d x

Apply Integration By Parts

: x ln ∣ 1 - x ∣ - \int - \frac{x}{1 - x} d x

= x ln ∣ 1 - x ∣ - \int - \frac{x}{1 - x} d x

\int - \frac{x}{1 - x} d x = - 1 + x + ln ∣ 1 - x ∣

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

Simplify

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

Add a constant to the solution

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

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What is the integral of ln|1-x| ?
The integral of ln|1-x| is xln|1-x|+1-x-ln|1-x|+C