Here the integral does not converge. In order for a function to be integrable, it must be continuous on its domain. Here, notice that the function \(\displaystyle{\frac{{{1}}}{{{\left({x}−{1}\right)}^{{{2}}}}}}\)

has a discontinuity at \(\displaystyle{x}={1}\in{\left[{0},{2}\right]}\). If we try to integrate beside the discontinuity, we also encounter problems with the improper integrals:

\(\displaystyle{\int_{{{0}}}^{{{2}}}}{\frac{{{1}}}{{{\left({x}-{1}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}={\int_{{{0}}}^{{{1}}}}{\frac{{{1}}}{{{\left({x}-{1}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}+{\int_{{{1}}}^{{{2}}}}{\frac{{{1}}}{{{\left({x}-{1}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}\)

Notice that: \(\displaystyle{\int_{{{0}}}^{{{1}}}}{\frac{{{1}}}{{{\left({x}-{1}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}=\lim{t}\rightarrow{1}{\int_{{{0}}}^{{{t}}}}{\frac{{{1}}}{{{\left({x}-{1}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}=\lim{t}\rightarrow{1}{\left({\frac{{-{t}}}{{{t}-{1}}}}\right)}\) which does not exist since we have this limit approach ∞ and −∞ from the left and right, respectively.

Therefore, we cannot integrate by this method either.

has a discontinuity at \(\displaystyle{x}={1}\in{\left[{0},{2}\right]}\). If we try to integrate beside the discontinuity, we also encounter problems with the improper integrals:

\(\displaystyle{\int_{{{0}}}^{{{2}}}}{\frac{{{1}}}{{{\left({x}-{1}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}={\int_{{{0}}}^{{{1}}}}{\frac{{{1}}}{{{\left({x}-{1}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}+{\int_{{{1}}}^{{{2}}}}{\frac{{{1}}}{{{\left({x}-{1}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}\)

Notice that: \(\displaystyle{\int_{{{0}}}^{{{1}}}}{\frac{{{1}}}{{{\left({x}-{1}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}=\lim{t}\rightarrow{1}{\int_{{{0}}}^{{{t}}}}{\frac{{{1}}}{{{\left({x}-{1}\right)}^{{{2}}}}}}{\left.{d}{x}\right.}=\lim{t}\rightarrow{1}{\left({\frac{{-{t}}}{{{t}-{1}}}}\right)}\) which does not exist since we have this limit approach ∞ and −∞ from the left and right, respectively.

Therefore, we cannot integrate by this method either.