因为a+[1/(e^-x-1)]=a+[e^x/(1-e^x)]【这一步是分式的分子分母同乘以e^x】而，已知a+[1/(e^-x+1)]=-{a+[1/(e^x-1)]所以：a+[e^x/(1-e^x)]=-a-[1/(e^x-1)]==> 2a=[e^x/(e^x-1)]-[1/(e^x-1)]==> 2a=[e^x-1]/(e^x-1)==> 2a=1==> a=1/2
