求一道不定积分 x^2/x^4+1
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∫x^2dx/(x^4+1)
=(1/2)∫xd(x^2)/[(x^2+1+√2x)(x^2+1-√2x)]
=(1/4√2)[ ∫d(x^2)/(x^2+1-√2x)-∫dx^2/(x^2+1+√2x)]
=(1/4√2) [ ∫d(x^2-√2x+1)/(x^2+1-√2x)+∫d√2x/(x^2+1-√2x)
-∫d(x^2+√2x+1)/(x^2+√2x+1)+∫d√2x/(x^2+√2x+1)]
=(1/4√2)ln[ |x^2+1-√2x|/|x^2+1+√2x|] +(1/4)[∫dx/((x-√2/2)^2+1/2)+∫dx/((x+√2/2)^2+1/2)]
=(1/4√2)ln|x^2+1-√2x|/|x^2+1+√2x|+(1/2√2)arctan(√2x-1)+(1/2√2)arctan(√2x+1)+C
=(1/2)∫xd(x^2)/[(x^2+1+√2x)(x^2+1-√2x)]
=(1/4√2)[ ∫d(x^2)/(x^2+1-√2x)-∫dx^2/(x^2+1+√2x)]
=(1/4√2) [ ∫d(x^2-√2x+1)/(x^2+1-√2x)+∫d√2x/(x^2+1-√2x)
-∫d(x^2+√2x+1)/(x^2+√2x+1)+∫d√2x/(x^2+√2x+1)]
=(1/4√2)ln[ |x^2+1-√2x|/|x^2+1+√2x|] +(1/4)[∫dx/((x-√2/2)^2+1/2)+∫dx/((x+√2/2)^2+1/2)]
=(1/4√2)ln|x^2+1-√2x|/|x^2+1+√2x|+(1/2√2)arctan(√2x-1)+(1/2√2)arctan(√2x+1)+C
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