Question

How do you evaluate the integral `int ln(x-1)` from 1 to 2?

Answer 1

Please see the explanation section below.

Explanation:

`int_1^2 ln(x-1) dx` is an improper integral because the integrand is not defined at one of the limits of integration. (`ln(x-1)` is not defined for `x=1`.)

Before attempting to evaluate this improper integral, let's change the variable of integration. (Let's do a substitution.)

Let `t = x-1`, so that `dt = dx` and when `x=1`, the `t = 0` and when `x=2`, then `t=1`

Our integral becomes

`int_0^1 lnt dt = lim_(ararr0) int_a^1 lnt dt` `" "` if the limit exists.

We need `int lnt dt` which may be found by integration by parts. with `u = lnt` and `vd = dt`

`int lnt dt = tlnt-t +C`

Returning to the main question,

`int_0^1 lnu du = lim_(ararr0^+) int_a^1 lnt dt`

`= lim_(ararr0^+) [t ln t -t]_a^1`

`= lim_(ararr0^+) [(0-1) - (a ln a -a)]`

`= -1 - lim_(ararr0^+) (a ln a-a)`

`= -1 - lim_(ararr0^+) (a ln a)`

To evaluate `lim_(ararr0^+) (a ln a)`, we'll use l"Hospital's Rule.

`lim_(ararr0^+) (a ln a)` has initial form `0*-oo`, so we need to rewrite

`= lim_(ararr0^+) (ln a/(1/a))` has form `(-oo)/oo` so we try

`= lim_(ararr0^+) ((1/a)/(-1/a^2)) = lim_(ararr0^+) (-a) = 0`.

So, `int_0^1 lnu du = -1 - lim_(ararr0^+) (a ln a) = -1`

And finally,

`int_1^2 ln(x-1) dx = -1` .