What is the integral of ln((x-1)^2) ?

The integral of ln((x-1)^2) is 2xln(x-1)-2(x-1+ln|x-1|)+C

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integral of ln((x-1)^2)

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

= x ln ((x - 1)^{2}) - 2 (x - 1 + ln ∣ x - 1 ∣)

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

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

Add a constant to the solution

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

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