Question

What is `int 1/ (x^2 -2x +1)dx`?

Answer 1

I found:
`int1/(x^2-2x+1)dx=-1/(x-1)+c`

Explanation:

We can write:
`int1/(x^2-2x+1)dx=int1/(x-1)^2dx=`
set:
`x-1=t`
`x=t+1`
so
`dx=dt`
substituting:
`=int1/t^2dt=intt^-2dt=-t^-1+c=-1/t+c`
where `t=x-1` so going back to `x`:
`int1/(x^2-2x+1)dx=-1/(x-1)+c`