# int_(1)^(oo) (ln(2x-1))/(x^2) dx Evaluate

Evaluate ${\int }_{1}^{\mathrm{\infty }}\frac{\mathrm{ln}\left(2x-1\right)}{{x}^{2}}$
My approach is to calc
${\int }_{1}^{X}\frac{\mathrm{ln}\left(2x-1\right)}{{x}^{2}}dx$
and then take the limit for the answer when $X\to \mathrm{\infty }$
However, I must do something wrong. The correct answer should be $2\mathrm{ln}\left(2\right)$
${\int }_{1}^{X}\frac{\mathrm{ln}\left(2x-1\right)}{{x}^{2}}dx={\left[-\frac{1}{x}\mathrm{ln}\left(2x-1\right)\right]}_{1}^{X}+{\int }_{1}^{X}\frac{1}{x}×\frac{2}{2x-1}dx=-\frac{1}{X}\mathrm{ln}\left(2X-1\right)+2{\int }_{1}^{X}-\frac{1}{x}+\frac{2}{2x-1}dx=-1\frac{1}{X}\mathrm{ln}\left(2X-1\right)-2\mathrm{ln}X+2\mathrm{ln}\left(2X-1\right)$
Am I wrong? If I'm not, how to proceed?
=== EDIT ===
After the edit I wonder if this is the correct way to proceed:
$-\frac{1}{X}\mathrm{ln}\left(2X-1\right)-2\mathrm{ln}X+2\mathrm{ln}\left(2X-1\right)$
The first part will do to zero because of $\frac{1}{X}$ so we ignore that one, the second and third part:
$-2\mathrm{ln}X+2\mathrm{ln}\left(2X-1\right)=2\mathrm{ln}\left(\frac{2X-1}{X}\right)=2\mathrm{ln}\left(2-\frac{1}{X}\right)=2\mathrm{ln}\left(2\right)$
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seguidora1e
Your derivative is incorrect, it should be
$\frac{2}{2x-1}$
$\frac{1}{2x-1}$
Everything else seems correct.
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Haiphongum
Another approach :
Setting $x↦\frac{1}{x}$, we will obtain

Note that

hence
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