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不定积分(x的平方加2x加2)分之Dx怎么做

2012-12-19 求不定积分 ∫[(2xe^-x^2)+(1/√x)(cos√... 2013-12-02 求不定积分∫√(x^2+a^2)dx,要详细过程。。。 8 2013-12-...

如图

原式=∫1/[(x+1)²+1]dx =arctan(x+1)+C

∫[ x^2/(x^2+2)] dx =∫[ 1 - 2/(x^2+2)] dx = x -2∫ dx/(x^2+2) let x= √2 tany dx =√2 (secy)^2 dy ∫ dx/(x^2+2) =(√2/2)∫ dy =(√2/2)y + C' =(√2/2)arctan(x/√2) + C' ∫[ x^2/(x^2+2)] dx = x -2∫ dx/(x^2+2) = x -√2arctan(x/√2) + C

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 计算不定积分x平方倍一加x平方分之二加x平方过程答:∫x^2/(1+x)dx=∫(x^2-1+1)dx/(1+x)=∫(x^2-1)dx/(x+1)+∫dx/(x+1)=∫(x-1)dx+ln|x+1|=x^2/2-x+ln|x+1| +C

解 ∫x/(x²+2x+2)dx =1/2∫(2x+2-2)/(x²+2x+2)dx =1/2∫(2x+2)/(x²+2x+2)dx-∫1/(x²+2x+2)dx =1/2∫1/(x²+2x+2)d(x²+2x+2)-∫1/[(x+1)²+1]dx =1/2∫1/udu-∫1/[(x+1)²+1]d(x+1) =1/2ln|u|-∫1/(u²+1)du =1/...

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