Question

# int_{0}^{2}frac{1}{(x-1)^{2}}dx

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

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. ​