求函数f(x)=∫(上限0,下限-1)(1+t)arctantdt的极小值

f(x) = ∫(-1->x)(1+t)arctantdt
f'(x) = (1+x)arctanx=0
x=-1 or x=0
f''(x) = (1+x)/(1+x^2) + arctanx
f''(0) >0 (min)
min f(x)
=f(0)
= ∫(-1->0)(1+t)arctantdt
=(1/2) ∫(-1->0) arctant d(1+t)^2
=(1/2) [(1+t)^2arctant](-1->0) - (1/2) ∫(-1->0) (1+t)^2/(1+t^2) dt
= - (1/2) [∫(-1->0) dt + ∫(-1->0) 2t/(1+t^2) dt ]
= -(1/2){ [t](-1->0) + [ln(1+t^2)](-1->0) }
=-(1/2) { 1+ln2 }