根据积分中值定理，存在c∈[0,1]，使得∫(0,1) {[ln(1+x)]^n}/(1+x^2)dx={[ln(1+c)]^n}/(1+c^2)原式=lim(n->∞) {[ln(1+c)]^n}/(1+c^2)因为0<=ln(1+c)<=ln2<1所以原式=0
