如题所述
â«[e,e^2] lntdt/(t-1)^2
= -â«[e,e^2] lntd(1/(t-1))
=-lnt/(t-1) |[e,e^2] +â«[e,e^2] (dt/t(t-1)
=-2/(e^2-1)+1/(e-1)+â«[e,e^2]dt/(t-1)-â«[e,e^2]dt/t
=-2/(e^2-1)+1/(e-1)+ln[(e^2-1)/(e-1)]-2+1
=-2/(e^2-1)+1/(e-1)+ln(e+1)-1
= -â«[e,e^2] lntd(1/(t-1))
=-lnt/(t-1) |[e,e^2] +â«[e,e^2] (dt/t(t-1)
=-2/(e^2-1)+1/(e-1)+â«[e,e^2]dt/(t-1)-â«[e,e^2]dt/t
=-2/(e^2-1)+1/(e-1)+ln[(e^2-1)/(e-1)]-2+1
=-2/(e^2-1)+1/(e-1)+ln(e+1)-1
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第1个回答 2012-05-06
................你教我吧
