${\displaystyle \underset{-1}{\overset{1}{}}}\frac{1}{(x-1{)}^2}dx=$

Select one:

$\underset{b{1}^{+}}{lim}{\displaystyle \underset{-1}{\overset{b}{}}}\frac{1}{(x-1{)}^2}dx$

$\underset{b-1}{lim}{\displaystyle \underset{b}{\overset{1}{}}}\frac{1}{(x-1{)}^2}dx$

$\underset{b-1^{+}}{lim}{\displaystyle \underset{b}{\overset{1}{}}}\frac{1}{(x-1{)}^2}dx$

None of these answers

$\underset{b1}{lim}{\displaystyle \underset{-1}{\overset{b}{}}}\frac{1}{(x-1{)}^2}dx$