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

Take the constant out:

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

= - \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

\int_{0}^{H} z^{2} (1 - z)^{2} d z

(x^{2} + y) + (x + e^{y}) y^{^{'}} = 0

\frac{d y}{d x} = \frac{cos ( 10 x )}{e ^{10 y}}

x \to 0 lim (\frac{2 ^{x}}{3 ^{x}})

\frac{\partial}{\partial x} (ln (3 x + 5 y))

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