※ 引述《hothero (hothero)》之銘言：
: ∫ln(x^2-1)dx
: 3Q~~
令 u = ln(x^2 - 1) , dv = dx
2x
則 du = --------- dx , v = x
x^2 - 1
∫ln(x^2 - 1) dx
2x
= (x)(ln(x^2 - 1)) - ∫(x)(---------) dx
x^2 - 1
(2)(x^2)
= (x)(ln(x^2 - 1)) - ∫---------- dx
x^2 - 1
2
= (x)(ln(x^2 - 1)) - ∫2 + ------- dx
x^2 - 1
1 1
= (x)(ln(x^2 - 1)) - ∫2 + ----- - ----- dx
x - 1 x + 1
= (x)(ln(x^2 - 1)) - (2x + ln|x - 1| - ln|x + 1|) + c
= (x)(ln(x^2 - 1)) - 2x - ln|x - 1| + ln|x + 1| + c
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