问题标题：
x的4次方加1的倒数的不定积分是?
问题描述：
x的4次方加1的倒数的不定积分是?
葛骅回答：
∫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√2)]arctan[(x-1/x)√2]-[1/(4√2)]ln|[(x+1/x-√2)/(x+1/x+√2)|+C
=[1/(2√2)]arctan[x/√2-1/(√2*x)]-[1/(4√2)]ln|(x²-√2*x+1)/(x²+√2*x+1)|+C
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