第1个回答 2017-04-07
∫1/(x⁴+1)dx=(1/2)∫[(x²+1)-(x²-1)]/(x⁴+1)dx=(1/2)∫(x²+1)/(x⁴+1)dx-(1/2)∫(x²-1)/(x⁴+1)dx=(1/2)∫(1+1/x²)/(x²+1/x²)dx-(1/2)∫(1-1/x²)/(x²+1/x²)dx=(1/2)∫d(x-1/x)/[(x-1/x)²+2]-(1/2)∫d(x+1/x)/[(x+1/x)²-2]=(1/2)(1/√2)arctan[(x-1/x)/√2]-(1/2)[1/(2√2)]ln|[(x+1/x)-√2]/[(x+1/x)+√2]|+C=(√2/4)arctan[(x-1/x)/√2]-(√2/8)ln|(x²-√2x+1)/(x²+√2x+1)|+C本回答被网友采纳
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