x=1 - особая точка
[m]u=ln(1-x)[/m]
[m]du=\frac{1}{1-x}\cdot (1-x)`dx[/m] ⇒ [m]du=\frac{-1}{1-x}dx[/m]
[m]\frac{-1}{1-x}dx=-d(ln(1-x))[/m]
[m]\int^{1}_{\frac{1}{2}}\frac{dx}{(1-x)ln^2(1-x)}=-lim_{a → 1-0}\int^{a}_{\frac{1}{2}}\frac{d(ln(1-x))}{ln^2(1-x)}=[/m]
[m]=-lim_{a → 1-0}\int^{a}_{\frac{1}{2}}ln^{-2}(1-x)d(ln(1-x))=[/m]
[m]=lim_{a → 1-0}ln^{-1}(1-x)|^{a}_{\frac{1}{2}}=
lim_{a → 1-0}\frac{1}{ln(1-x)}|^{1}_{\frac{1}{2}}=[/m]
[m]=lim_{a → 1-0}\frac{1}{ln(1-a)}-\frac{1}{ln(1-\frac{1}{2})}=0-\frac{1}{ln\frac{1}{2}}=\frac{1}{ln2}[/m]