Integral with logarithm in denominator

Question

I am having hard time to find following integral:

\begin{equation} \int\frac{2x-1}{2x(1-x)}\cdot\frac{1}{2(1-x)+\log{x}}dx \end{equation}

Can anyone help? Thank you.

Answer 1

This is just a guess. I have not worked it out. Here is a possible way to start:

If $w = 2(1-x)+\log x$ then $dw = \left(-2+\dfrac{1}{x}\right)dx = \dfrac{1-2x}{x}dx$

The first term is $\dfrac{-dw}{2(1-x)}$

So, let $u = \dfrac{1}{2(1-x)}$ and $dv = \dfrac{-dw}{w}$. Try integration by parts.