求一个定积分的题目，谢谢

Question

如题所述

Answer 1

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

Answer 2

∫ (0->1) ln(1+x)/(2-x)^2 dx
=∫ (0->1) ln(1+x) d [1/(2-x)]
=[ln(1+x)/(2-x)](0->1) -∫ (0->1) dx/(1+x)(2-x)
=2ln2 - (1/3) ∫ (0->1)[ 1/(1+x)+1/(2-x)] dx
=2ln2 -(1/3)[ ln[(1+x)/(2-x)] ] (0->1)
=2ln2 -(1/3) (2ln2)
= (4/3)ln2