Question

How do you solve `int1/(y^2 - 2y + 1)dy`?

Answer 1

`= 1/(1-y) + C`

Explanation:

`int \ 1/(y^2 - 2y + 1 ) \ dy`

`int \ 1/(y-1)^2 \ dy`

by power rule

`=- 1/(y-1) + C`

tidy up
`= 1/(1-y) + C`