推 stifler218:非常感謝您的解答！感謝！ 134.208.44.21 02/25 13:44
※ 引述《stifler218 (我愛朝比奈)》之銘言：
: ∫1/x‧ln|x|‧ln|ln|x|| dx
: 感謝
set t=ln|x| , dt=dx/x
∫1/x‧ln|x|‧ln|ln|x|| dx = ∫t‧ln|t| dt
set u=ln|t| , dv=tdt
du=dt/t , v=t^2/2
integration by part
=> ∫t‧ln|t| dt = (t^2/2)‧ln|t| - (1/2)∫tdt
= (t^2/2)‧ln|t| - (1/4)‧t^2 + C
= (1/2)‧(ln|x|)^2 ‧ln|ln|x|| - (1/4)‧(ln|x|)^2 + C
so
∫1/x‧ln|x|‧ln|ln|x|| dx = (1/2)‧(ln|x|)^2‧ln|ln|x|| - (1/4)‧(ln|x|)^2 + C
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