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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{ln ( u )}{u ^{2}} d u

= \int \frac{ln ( u )}{u ^{2}} d u

Apply Integration By Parts

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

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

\int - \frac{1}{u ^{2}} d u = \frac{1}{u}

= - \frac{ln ( u )}{u} - \frac{1}{u}

Substitute back

u = x - 1

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

Add a constant to the solution

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

\int \frac{25}{x ^{\frac{1}{2}}} d x

(csc (x) - cot (x))^{- 1}

f (x) = \frac{1}{x ^{4}}, (1, 1)

\int sinh (9 x) d x

\int \frac{ln ( 18 x )}{x} d x

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