求不定积分 ∫5/(x²+4x+9)dx
人气:215 ℃ 时间:2020-04-10 02:36:08
解答
x^2+4x+9 = (x+2)^2+5
let
x+2 = √5tana
dx =√5(seca)^2 da
∫5/(x²+4x+9)dx
=∫[5/(5(seca)^2)] √5(seca)^2 da
=√5a+ C
=√5arctan[(x+2)/√5] + C
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