Question

$\underset{0}{\overset{1}{\int }}\log \left(\frac{1}{x}-1\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^1\log \left(\frac{1}{x}-1\right)dx...\left(\mathrm{i}\right) \\ ={\int }_0^1\log \left(\frac{1}{1-x}-1\right)dx\left[\mathrm{Using}{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx\right] \\ I={\int }_0^1\log \left(\frac{x}{1-x}\right)dx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ 2I={\int }_0^1\log \left(\frac{1-x}{x}\right)+\log \left(\frac{x}{1-x}\right)dx \\ ={\int }_0^1\log \left(\frac{1-x}{x}\times \frac{x}{1-x}\right)dx \\ ={\int }_0^1\log 1dx \\ =0 \\ \mathrm{Hence}I=0 \end{gathered}$$