Question

How do you integrate of `1/(1-x)`?

Answer 1

`int (dx)/(1-x) = -lnabs(1-x)`

Explanation:

Substitute:

`t= 1-x`
`dt = -dx`

and you have:

`int (dx)/(1-x) = -int (dt)/t = -ln abs (t) = -lnabs(1-x)`