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1/(4x^2+4x+5)的不定积分怎么求的啊?
人气:113 ℃ 时间:2020-06-15 20:19:40
解答
∫dx/(4x^2+4x+5)=
=∫dx/[(2x+1)²+2²]=
=(1/4)∫dx/[(x+1/2)²+1]=
=(1/4)∫d(x+1/2)/[(x+1/2)²+1]=
=(1/4)arctan(x+1/2)+c
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