∫dt/2t(1+t^2) =∫[1/2t(1+t^2)]dt =∫[1/t[(t^2)/2+1/2] dt查积分表，有：∫dx/[x(ax^2+b)=(1/2b)ln(x^2/|ax^2+b|)+C对比题目，可知：x=t、a=1/2、b=1/2所以：∫[1/t[(t^2)/2+1/2] dt=1/4ln[t^2/|(1/2)t^2+1/2|]+C
原式=(1/2)∫[1/t-t/(t^2+1)]dt=(1/2)ln|t|-(1/2)*(1/2)∫d(t^2+1)/(t^2+1)=(1/2)ln|t|-(1/4)ln(1+t^2)+C.
∫dt/(2t(t^2+1))=(1/2)∫ ( 1/t - t/(t^2+1) dt=(1/2)[ ln|t| + (1/2)ln|t^2+1| ] +C
设tanu=t,1+t²=sec²u,dt=sec²udu,sinu=t/√(1+t²)原式=∫sec²u/(2tanu*sec²u)du=1/2∫cotudu=1/2*ln|sinu|+C=1/2ln|t/√(1+t²)|+C