This discussion is now closed.
\dfrac{1}{y}\mbox{d}y
\displaystyle \int \dfrac{1}{y}\ \mbox{d}y = \int \dfrac{2x - 1}{(x-1)(2x-3)}\ \mbox{d}x
\displaystyle \int \dfrac{2x - 1}{(x-1)(2x-3)}\ \mbox{d}x = \int \left( \dfrac{-1}{x - 1} + \dfrac{4}{2x - 3} \right) \mbox{d}x
\newline \displaystyle \int \dfrac{1}{y}\ \dfrac{\mbox{d}y}{\mbox{d}x}\ \mbox{d}x = \int \left( \dfrac{-1}{x-1} + \dfrac{4}{2x - 3}\ \right) \mbox{d}x
\newline \displaystyle \int \dfrac{1}{y}\ \mbox{d}y = \int \left( \dfrac{-1}{x-1} + \dfrac{4}{2x - 3}\ \right) \mbox{d}x