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

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

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integral of (ln(x-1))/(x-1)

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

\frac{ln ^{2} ( x - 1 )}{2} + C

Solution steps

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

Apply u-substitution

: \int \frac{ln ( u )}{u} d u

= \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{v ^{2}}{2}

Substitute back

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

Add a constant to the solution

= \frac{ln ^{2} ( x - 1 )}{2} + C

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