\(\int\limits_0^1\frac1{[1-log_2(1-x)]}dx\)
= \(\int\limits_0^1\cfrac1{[log_2\frac2{(1-x)}]}dx\) (\(\because log_22=1 \& log\,A- log\,B = log\frac{A}{B}\))
\(\therefore\) 0 < x < 1
⇒ 0 < 1 - x < 1
⇒ 2/(1-x) > 2
Now,
\(\int\limits_0^1\cfrac1{[log_2\frac2{1-x}]}dx\) = ....+ \(\int\limits_{15/16}^{7/8}\cfrac1{[log_2\frac2{1-x}]}dx\) + \(\int\limits_{7/8}^{3/4}\cfrac1{[log_2\frac2{1-x}]}dx\) + \(\int\limits_{3/4}^{1/2}\cfrac1{[log_2\frac2{1-x}]}dx\) + \(\int\limits_{1/2}^{1}\cfrac1{[log_2\frac2{1-x}]}dx\)
= .... + \(\int\limits_{15/16}^{7/8}\frac1{4}dx\) + \(\int\limits_{7/8}^{3/4}\frac1{3}dx\) + \(\int\limits_{3/4}^{1/2}\cfrac1{2}dx\)
= ....+ \(\frac14(\frac{14}{16}-\frac{15}{16})+\frac13(\frac68-\frac78)\) + \(\frac12(\frac24-\frac34)+\frac12(1-\frac12)\)
= \(\frac14-\frac18-\frac1{24}-\frac1{64}-..... <4\)
Hence, \(\int\limits_0^1\frac1{1-log_2(1-x)}dx<1/4\)
