第2个回答 2010-04-11
解:l=∫(√ 3, 2)√1+y'^2 dx
y'=1/x
=>l=∫(√ 3, 2)√1+1/x^2 dx
=∫(√ 3, 2)x*√1+x^2/x^2 dx
=1/2∫(√ 3, 2)√1+x^2/x^2 dx^2 -----(1)
令1+x^2=t^2(t>0)=>√1+x^2=t 且2<t<√5d(x^2)=d(t^2-1)=2tdt
=>(1)=1/2∫(2,√5)t/(t^2-1)2tdt=∫(2,√5)t^2/(t^2-1)dt
=∫(2,√5)dt+∫(2,√5)1/(t^2-1)dt
=√5-2+1/2∫(2,√5)d(ln((t-1)/(t+1))
=√5-2+ln((√15-√3)/2)