像这种积分问题,令 x = tant 会更好解决：
令 x = tant,则 dx = (sect)^2*dt
∫[x^2/(1+x^2)]*arctanx*dx
=∫[(tant)^2/(sect)^2] *t *(sect)^2*dt 注：1+(tant)^2 = (sect)^2
=∫t*(tant)^2*dt
=∫t*[(sect)^2 - 1]*dt
=∫t*(sect)^2 *dt - ∫t*dt
=t*(tant) - ∫(tant)*dt - 1/2*t^2
=t*tant -1/2*t^2 - ∫sint*dt/cost
=t*tant - 1/2*t^2 + ∫d(cost)/cost
=t*tant - 1/2*t^2 + ln|cost| + C
=x*arctanx - 1/2*(arctanx)^2 + 1/2*ln(cost)^2 + C
=x*arctanx - 1/2*(arctanx)^2 - 1/2*ln(sect)^2 + C
=x*arctanx - 1/2*(arctanx)^2 - 1/2*ln(1+x^2) + C