Question

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

Answer 1

Answer: 0

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