Question

$\int\limits_{1}^{\infty} \frac{\ln (x-1) d x}{x^{2} \ln x \cdot \ln \left(\frac{x}{x-1}\right)}$ is equal to -

Answer 1

Put $x=\frac{1}{t} \Rightarrow d x=-\frac{1}{t^{2}} d t$
Given integral
$I=\int\limits_{0}^{1} \frac{\ln \left(\frac{1}{t}-1\right) d t}{\ln t \cdot \ln (1-t)}=\int\limits_{0}^{1}\left(\frac{1}{\ln t}-\frac{1}{\ln (1-t)}\right) d t$
$=\int\limits_{0}^{1} \frac{d t}{\ln t}-\int\limits_{0}^{1} \frac{d t}{\ln (1-t)}=\int\limits_{0}^{1} \frac{d t}{\ln (1-t)}-\int\limits_{0}^{1} \frac{d t}{\ln (1-t)}=0$