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: 
$$ -2\ln X+2\ln(2X-1) = 2\ln \left( \frac{2X-1}{X}\right) = 2\ln \left( 2-\frac{1}{X}  \right) = 2\ln (2)$$

Your derivative is incorrect, it should be $$\frac{2}{2x-1}$$ instead of $$\frac{1}{2x-1}$$ Everything else seems correct.
