Evaluate $\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2} $ $$\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2} dx$$
My approach is to calc
$$\int_{1}^{X} \frac{\ln{(2x-1)}}{x^2} dx$$ and then take the limit for the answer when $X \rightarrow \infty$
However, I must do something wrong. The correct answer should be $2\ln(2)$.
$$\int_{1}^{X} \frac{\ln{(2x-1)}}{x^2} dx = \left[-\frac{1}{x} \ln (2x-1) \right]_{1}^{X} + \int_{1}^{X} \frac{1}{x} \times \frac{2}{2x-1} dx = -\frac{1}{X}\ln(2X-1) + 2\int_{1}^{X} -\frac{1}{x} + \frac{2}{2x-1} dx = -1\frac{1}{X}\ln(2X-1)-2\ln X+2\ln(2X-1) $$
Am I wrong? If I'm not, how to proceed? 
=== EDIT ===
After the edit I wonder if this is the correct way to proceed:
$$ - \frac{1}{X}\ln(2X-1)-2\ln X+2\ln(2X-1) $$ The first part will do to zero because of $\frac{1}{X} $ so we ignore that one, the second and third part: 
$$ -2\ln X+2\ln(2X-1) = 2\ln \left( \frac{2X-1}{X}\right) = 2\ln \left( 2-\frac{1}{X}  \right) = 2\ln (2)$$
 A: Your derivative is incorrect, it should be $$\frac{2}{2x-1}$$ instead of $$\frac{1}{2x-1}$$ Everything else seems correct.
A: Another approach :
Setting $x\mapsto\frac1x$, we will obtain
$$
\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2}\ dx=\int_{0}^{1} \ln\left(\frac{2-x}{x}\right) dx=\int_{0}^{1} \bigg[\ln(2-x)-\ln x\bigg]\ dx.
$$
Note that
$$
\int\ln y\ dy=y\ \ln y-y+C,
$$
hence
$$
\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2}\ dx=2\ln 2.
$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{1}^{\infty}{\ln\pars{2x - 1} \over x^{2}}\,\dd x:\ {\large ?}}$

\begin{align}
&\color{#66f}{\large\int_{1}^{\infty}{\ln\pars{2x - 1} \over x^{2}}\,\dd x}
=-\int_{x\ =\ 1}^{x\ \to\infty}\ln\pars{2x - 1}\,\dd\pars{1 \over x}
=\int_{1}^{\infty}{1 \over x}\,{2 \over 2x - 1}\,\dd x
\\[3mm]&
=\int_{1}^{\infty}{1 \over x\pars{x - 1/2}}\,\dd x
=2\int_{1}^{\infty}\pars{{1 \over x - 1/2} - {1 \over x}}\,\dd x
\\[3mm]&=\left. 2\ln\pars{\verts{x - 1/2 \over x}}\right\vert_{1}^{\infty}
=\color{#66f}{\large 2\ln\pars{2}} \approx {\tt 1.3863}
\end{align}

