※ 引述《TaiwanBank (澳仔金控台灣分行)》之銘言：
: 1
: Show that ∫ (x-1)/lnx dx = ln2.
: 0
1 1
令 I(y) = ∫ (x^y -1)/lnx dx => I'(y) = ∫ x^y dx = 1/(y+1)
0 0
故 I(y) = ln(y+1) + C, 但 I(0) = 0 = 0 + C => C = 0
I(1) = ln(2)
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