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2018-09-21 · 知道合伙人教育行家
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令 x=t^2,则 dx=2tdt,
原式 = ∫2tdt/[t(1+t^2)]
= 2∫dt/(1+t^2)
= 2arctant + C
= 2arctan√x + C
原式 = ∫2tdt/[t(1+t^2)]
= 2∫dt/(1+t^2)
= 2arctant + C
= 2arctan√x + C
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