作者fsshman (翼德)
看板trans_math
标题[积分] 求助!
时间Tue Oct 27 00:41:27 2009
r
∫----------------------------------- dr
[ r^2 + h^2 - rh*2^(1/2) ]^(1/2)
答案是
[h^2+r^2-rh*2^(1/2)]^(1/2)
+h/2^(1/2)*ln[2(h^2+r^2-rh*2^(1/2))^(1/2)+2r-h*2^(1/2)]
我用代数变换去算(u=r-h/2^(1/2))，可是不知道为甚麽，ln[]的里面少了两倍
[h^2+r^2-rh*2^(1/2)]^(1/2)
+h/2^(1/2)*ln[(h^2+r^2-rh*2^(1/2))^(1/2)+r-h/2^(1/2)]
屡试不爽，希望有人为我解惑一下，谢谢!
(跟号2打起来好麻烦= =)
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