用分部积分,设u=arctanx,v'=1/x^2u'=1/(1+x^2),v=-1/x,原式=-(arctanx)/x+∫ dx/[x(1+x^2)]=-(arctanx)/x+∫(-x) dx/(1+x^2)+∫ dx/x=-(arctanx)/x-(1/2)∫d(1+x^2)/(1+x^2)+∫ dx/x=-(arctanx)/x-(1/2)ln(1+x^2)+ln...