计算定积分∫(1~-0)ln(1+x)/(2-x)^2.dx

上下限看不清楚,先做不定积分吧
∫ ln(1 + x)/(2 - x)² dx
= - ∫ ln(1 + x)/(2 - x)² d(2 - x)
= ∫ ln(1 + x) d[1/(2 - x)]
= [ln(1 + x)]/(2 - x) - ∫ 1/(2 - x) · d[ln(1 + x)],分部积分法
= [ln(1 + x)]/(2 - x) - ∫ 1/[(2 - x)(1 + x)] dx
= [ln(1 + x)]/(2 - x) - (1/3)∫ [(2 - x) + (1 + x)]/[(2 - x)(1 + x)] dx
= [ln(1 + x)]/(2 - x) - (1/3)∫ [1/(1 + x) + 1/(2 - x)] dx
= [ln(1 + x)]/(2 - x) - (1/3)[ln|1 + x| - ln|2 - x|] + C
= [ln(1 + x)]/(2 - x) - (1/3)ln| (1 + x)/(2 - x) | + C
若上限是1,下限是0,则定积分
∫(0→1) ln(1 + x)/(2 - x)² dx
= [ln(1 + 1)]/(2 - 1) - (1/3)ln[ (1 + 1)/(2 - 1) ] - { [ln(1 + 0)]/(2 - 0) - (1/3)ln[ (1 + 0)/(2 - 0) ] }
= (1/3)ln(2)