What is the integral of log_{6}(x-1) ?

The integral of log_{6}(x-1) is xlog_{6}(x-1)-1/(ln(6))(x-1+ln|x-1|)+C

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integral of log_{6}(x-1)

\int lo g_{6} (x - 1) d x

x lo g_{6} (x - 1) - \frac{1}{ln ( 6 )} (x - 1 + ln ∣ x - 1 ∣) + C

Solution steps

\int lo g_{6} (x - 1) d x

Apply Integration By Parts

: x lo g_{6} (x - 1) - \int \frac{x}{ln ( 6 ) ( x - 1 )} d x

= x lo g_{6} (x - 1) - \int \frac{x}{ln ( 6 ) ( x - 1 )} d x

\int \frac{x}{ln ( 6 ) ( x - 1 )} d x = \frac{1}{ln ( 6 )} (x - 1 + ln ∣ x - 1 ∣)

= x lo g_{6} (x - 1) - \frac{1}{ln ( 6 )} (x - 1 + ln ∣ x - 1 ∣)

Add a constant to the solution

= x lo g_{6} (x - 1) - \frac{1}{ln ( 6 )} (x - 1 + ln ∣ x - 1 ∣) + C

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What is the integral of log_{6}(x-1) ?
The integral of log_{6}(x-1) is xlog_{6}(x-1)-1/(ln(6))(x-1+ln|x-1|)+C