3个回答
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解法一:
∫1/(1+X^4)dx
=∫1/(x^2-√2x+1)(x^2+√2x+1)dx
=√2/2{∫[(2x-√2+√2)/(x^2-√2x+1)-∫(2x+√2-√2)/(x^2+√2x+1)]dx }
解法二:
∫1/(1+x^4)dx
=∫[ (1/x^2)/(1/x^2 +x^2)dx
=1/2∫[ (1/x^2 +1) +(1/x^2 -1)]/(1/x^2 +x^2)dx
=1/2 [ ∫[ d(x-1/x)/[(x-1/x)^2 +2] -∫[ d(x+1/x)/[(x+1/x)^2 -2)]
∫1/(1+X^4)dx
=∫1/(x^2-√2x+1)(x^2+√2x+1)dx
=√2/2{∫[(2x-√2+√2)/(x^2-√2x+1)-∫(2x+√2-√2)/(x^2+√2x+1)]dx }
解法二:
∫1/(1+x^4)dx
=∫[ (1/x^2)/(1/x^2 +x^2)dx
=1/2∫[ (1/x^2 +1) +(1/x^2 -1)]/(1/x^2 +x^2)dx
=1/2 [ ∫[ d(x-1/x)/[(x-1/x)^2 +2] -∫[ d(x+1/x)/[(x+1/x)^2 -2)]
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