Evaluate ${\int}_{1}^{\mathrm{\infty}}\frac{\mathrm{ln}(2x-1)}{{x}^{2}}$

My approach is to calc

${\int}_{1}^{X}\frac{\mathrm{ln}(2x-1)}{{x}^{2}}dx$

and then take the limit for the answer when $X\to \mathrm{\infty}$

However, I must do something wrong. The correct answer should be $2\mathrm{ln}(2)$

${\int}_{1}^{X}\frac{\mathrm{ln}(2x-1)}{{x}^{2}}dx={[-\frac{1}{x}\mathrm{ln}(2x-1)]}_{1}^{X}+{\int}_{1}^{X}\frac{1}{x}\times \frac{2}{2x-1}dx=-\frac{1}{X}\mathrm{ln}(2X-1)+2{\int}_{1}^{X}-\frac{1}{x}+\frac{2}{2x-1}dx=-1\frac{1}{X}\mathrm{ln}(2X-1)-2\mathrm{ln}X+2\mathrm{ln}(2X-1)$

Am I wrong? If I'm not, how to proceed?

=== EDIT ===

After the edit I wonder if this is the correct way to proceed:

$-\frac{1}{X}\mathrm{ln}(2X-1)-2\mathrm{ln}X+2\mathrm{ln}(2X-1)$

The first part will do to zero because of $\frac{1}{X}$ so we ignore that one, the second and third part:

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