∫[(1+t)/(1+t^2)一1/(1+t)]dt=∫dt/(1+t^2)+∫tdt/(1+t^2)一∫dt/(1+t)=∫dt/(1+t^2)+(1/2)∫dt^2/(1+t^2)-∫dt/(1+t)=arctant+(1/2)ln(1+t^2)-ln(1+t)=arctant+(1/2)ln[(1+t^2)/(1+t)^2]
