∫x*ln(1+x^2)dx
=1/2积分:ln(1+x^2)d(1+x^2)
令1+x^2=t
=1/2积分:lntdt
=1/2[tlnt-积分:td(lnt)]
=1/2[tlnt-积分:dt]
=1/2[tlnt-t]+C
=1/2(1+x^2)ln(1+x^2)-(1+x^2)/2+C
(C 为常数)