解ï¼
令ln(1+x)=tï¼å1+x=e^t
xï¼0â1ï¼åtï¼0âln2
â«[0ï¼1][ln(1+x)/(2-x)²]dx
=â«[0ï¼1] ln(1+x)/[3-(1+x)]² d(1+x)
=â«[0ï¼ln2] [t/(3-e^t)²]d(e^t)
=â«[0ï¼ln2] [t·e^t/(3-e^t)²]dt
=â«[0ï¼ln2] td[1/(3-e^t)]
=t/(3-e^t)|[0ï¼ln2] -â«[0ï¼ln2] [1/(3-e^t)]dt
=ln2/[3-e^(ln2)]-0/(3-1) -⅓â«[0ï¼ln2] [(3-e^t+e^t)/(3-e^t)]dt
=ln2/(3-2) -0 -⅓â«[0ï¼ln2][1 + e^t/(3-e^t)]dt
=ln2 -⅓â«[0ï¼ln2]dt +⅓â«[0ï¼ln2][1/(3-e^t)]d(3-e^t)
=ln2 -⅓t|[0ï¼ln2]+⅓ln|3-e^t||[0ï¼ln2]
=ln2-⅓(ln2 -0) +⅓[ln|3-e^(ln2)| -ln|3-1|]
=ln2-⅓ln2 +⅓[ln(3-2) -ln2]
=ln2-⅓ln2+⅓(0 -ln2)
=ln2-⅓ln2-⅓ln2
=⅓ln2
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