计算∫x*ln(1+x^2)dx=

∫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)

∫x*ln(1+x^2)dx
=1/2*∫ln(1+x^2)dx^2
令t=x^2。则
=1/2*∫ln(1+t)dt
=1/2*t*ln(1+t)-1/2*∫t/(1+t)dt
=1/2*t*ln(1+t)-1/2*∫[1-1/(1+t)]dt
=1/2*t*ln(1+t)-1/2*[t-ln(1+t)]+C
=1/2*t*ln(1+t)-1/2*t+1/2*ln(1+t)+C
代换回去得
1/2(1+x^2)ln(1+x^2)-(1+x^2)/2+C

(1+x^2)ln(1+x^2)－(1+x^2)+c