∫[0,1] [ln(1+x)]/[(2-x)^2]dx
= ∫[0,1] [ln(1+x)]d/[1/(2-x)]
= [0,1]|[ln(1+x) /(2-x)] - ∫[0,1][1/[(1+x)(2-x)]*dx
= ln2 - 1/3*∫[0,1][1/(1+x) + 1/(2-x)]*dx
=ln2 -1/3* [0,1]|[ln(1+x) + ln(2-x)]
=ln2