1 原式=x^4-8x^3+24x^2-24x+16
2 原式=x^4-10x^3+35x^2-50x+24

令x-1=t
dx=dt
∫2 0 f(x-1)dx
=∫1 -1 f(t)dt
=∫0 -1 f(t)dt + ∫1 0 f(t)dt
=∫0 -1 dt/(1+e^t) + ∫1 0 dt/(1+t)
=[t-ln(1+e^t)]0 -1 + [ln(1+x)]1 0
=1+ln(1+1/e)+ln2
=1+ln(2+2/e)

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)