第1个回答 2017-06-09
积分（1,2）1/x(x+1)dx
=积分(1,2)[1/x-1/(x+1)]dx
=[lnx-ln(x+1)],(1,2)
=[ln2-ln3]-[ln1-ln2]
=ln2-ln3+ln2
=2ln2-ln3
=ln4/3
=积分(1,2)[1/x-1/(x+1)]dx
=[lnx-ln(x+1)],(1,2)
=[ln2-ln3]-[ln1-ln2]
=ln2-ln3+ln2
=2ln2-ln3
=ln4/3
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