求下列不定积分:∫(1+2x)/[x(x+1)]dx 和∫1/(X²-X-6)dx
人气：150 ℃ 时间：2019-10-11 18:59:27
解答
令(1 + 2x)/[x(x + 1)] = A/x + B/(x + 1)令x = 0,A = (1 + 0)/(0 + 1) = 1令x = - 1,B = (1 - 2)/(- 1) = 1∴(1 + 2x)/[x(x + 1)] = 1/x + 1/(x + 1)∫ (1 + 2x)/[x(x + 1)] dx= ∫ 1/x dx + ∫ 1/(1 + x) dx= ln| ...
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