第1个回答 2008-06-30
设t=根号(x+1)
x=t^2-1
dx=2tdt
∫dx/[x*根号下(1+x)]
=∫2tdt/t(t^2-1)
=2∫dt/(t+1)(t-1)
=∫[1/(t-1)-1/(t+1)]dt
=ln|t-1|-ln|t+1|+C
=ln|(t-1)/(t+1)|+C
再把t用根号(x+1)带回本回答被提问者采纳
x=t^2-1
dx=2tdt
∫dx/[x*根号下(1+x)]
=∫2tdt/t(t^2-1)
=2∫dt/(t+1)(t-1)
=∫[1/(t-1)-1/(t+1)]dt
=ln|t-1|-ln|t+1|+C
=ln|(t-1)/(t+1)|+C
再把t用根号(x+1)带回本回答被提问者采纳
第2个回答 2008-06-30
分部积分
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