∫[1,0] dx/(x²-2x-3)
= ∫[1,0] dx/[(x-3)(x+1)]
= (1/4)∫[1,0] [1/(x-3) - 1/(x+1)] dx
= (1/4)ln|(x-3)/(x+1)|
= (1/4)ln(2/2) - (1/4)ln(3/1)
= -1/4ln3你這個(gè)好像是不定積分的求法?。?div style="margin-top:20px">
求定積分 ∫上1下01/(x^2-2x-3)dx
求定積分 ∫上1下01/(x^2-2x-3)dx
數(shù)學(xué)人氣：236 ℃時(shí)間：2020-04-11 04:05:14
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