计算：定积分∫（1 ,0）x/1+x^2 ..

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