※ 引述《tiffany9802 (Tiffany)》之銘言：
: 如題
: e
: S lnxdx
: 1
e
∫ lnx dx
1
|e e 1
= (x)(lnx) | - ∫ (x)(---) dx
|1 1 x
1
(令 u = lnx , dv = dx , 則 du = --- dx , v = x)
x
e
= (e)(ln(e)) - (1)(ln(1)) - ∫ 1 dx
1
|e
= (e)(1) - (1)(0) - (x) |
|1
= e - (e - 1) = e - e + 1 = 1
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