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

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

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integral of ln|x-1|

\int ln ∣ x - 1 ∣ d x

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

Solution steps

\int ln ∣ x - 1 ∣ d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

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

f (x) = \frac{2500 x}{ln ( 8 x + 8 )}

x y^{^{'}} + (x + \frac{1}{3}) y = e^{- 2 x} y^{- 2}

f (x) = x^{2} - x + 4, at x = 1

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