3.76: Partial fractions integration

Show that $\dfrac{1}{2x + 1} - \dfrac{1}{(2x + 1)^{2}} + \dfrac{4}{4x - 1} = \dfrac{24x^{2} + 14x + 4}{(2x + 1)^{2}(4x - 1)}$.

Hence find $\displaystyle\int_{\frac{1}{2}}^{1}\dfrac{24x^{2} + 14x + 4}{(2x + 1)^{2}(4x - 1)}\,\mathrm{d}x$, giving your answer in the form $\dfrac{1}{2}\mathrm{ln}\,p + q$, where $p$ and $q$ are rational numbers.