※ 引述《chris100466 (chris)》之銘言：
: 想請教要如何從這一步積分到這一步?
: http://i.imgur.com/JmGbDrC.jpg
y(cotα - c cotβ) + ch cotβ
------------------------------ = dx/dy
ch + (1 - c)y
cotα - c cotβ y ch ch cotβ
x = ----------------∫[1 - --------------] du + -------ln[[ch + (1 - c)y]/(ch)]
(1 - c) 0 ch + (1 - c)u (1 - c)
cotα - c cotβ ch ln[[ch + (1 - c)y]/(ch)]
= -----------------[y - ----------------------------]
(1 - c) (1 - c)
ch cotβ
+ -------ln[[ch + (1 - c)y]/(ch)]
1 - c
cotα - c cotβ ch(cotα - cotβ)
= ---------------- y + ------------------ ln[(ch)/[ch + (1 - c)y]]
(1 - c)^2 (1 - c)^2
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