Determine whether $\int_{0}^{2}\dfrac{1}{(x-1)^{2}}\,dx$ converges or diverges.

Solution Since $f(x)=\dfrac{1}{(x-1) ^{2}}$ is not defined at 1, the integral $\int_{0}^{2}\dfrac{1}{(x-1)^{2}}\,dx$ is an improper integral on the interval [ 0, 2]. We write the integral as follows: $$\int_{0}^{2}\dfrac{1}{(x-1)^{2}}\,dx=\int_{0}^{1}\dfrac{1}{(x-1)^{2}} \,dx+\int_{1}^{2}\dfrac{1}{(x-1)^{2}}\,dx$$

and investigate each of the two improper integrals on the right. $$\int_{0}^{1}\!\!\dfrac{1}{(x-1)^{2}}\,dx:\quad \lim_{t\,\rightarrow \,1^{-}}\int_{0}^{t}\dfrac{1}{(x-1)^{2}}\,dx=\lim_{t\rightarrow 1^{-}}\left[ \dfrac{-1}{x-1}\right] _{0}^{t}\!=\lim_{t\rightarrow 1^{-}}\left( \dfrac{-1}{t-1} -1\!\right)\! =\infty$$ $\int_{0}^{1}\dfrac{1}{(x-1) ^{2}}\,dx$ diverges, so there is no need to investigate the second integral.

The improper integral $\int_{0}^{2}\dfrac{1}{(x-1) ^{2}}\,dx$ diverges.