Answer
$${\text{The integral diverges}}{\text{}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{1}{{2x - 1}}} dx \cr
& {\text{The integrand is not continuous at }}x = \frac{1}{2},{\text{ then}} \cr
& \int_0^1 {\frac{1}{{2x - 1}}} dx = \int_0^{1/2} {\frac{1}{{2x - 1}}} dx + \int_{1/2}^1 {\frac{1}{{2x - 1}}} dx \cr
& = \mathop {\lim }\limits_{a \to {{\frac{1}{2}}^ - }} \int_0^a {\frac{1}{{2x - 1}}} dx + \mathop {\lim }\limits_{b \to {{\frac{1}{2}}^ + }} \int_b^1 {\frac{1}{{2x - 1}}} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to {{\frac{1}{2}}^ - }} \left[ {\ln \left| {2x - 1} \right|} \right]_0^a + \frac{1}{2}\mathop {\lim }\limits_{b \to {{\frac{1}{2}}^ + }} \left[ {\ln \left| {2x - 1} \right|} \right]_b^1 \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to {{\frac{1}{2}}^ - }} \left[ {\ln \left| {2a - 1} \right|} \right] + \frac{1}{2}\mathop {\lim }\limits_{b \to {{\frac{1}{2}}^ + }} \left[ { - \ln \left| {2b - 1} \right|} \right] \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to {{\frac{1}{2}}^ - }} \left[ {\ln \left| {2a - 1} \right|} \right] - \frac{1}{2}\mathop {\lim }\limits_{b \to {{\frac{1}{2}}^ + }} \left[ {\ln \left| {2b - 1} \right|} \right] \cr
& {\text{Evaluate the limits}} \cr
& = \frac{1}{2}\left[ {\ln \left| {2\left( {\frac{1}{2}} \right) - 1} \right|} \right] - \frac{1}{2}\mathop {\lim }\limits_{b \to {{\frac{1}{2}}^ + }} \left[ {\ln \left| {2\left( {\frac{1}{2}} \right) - 1} \right|} \right] \cr
& = \infty \cr
& {\text{The integral diverges}}{\text{.}} \cr} $$
