1/(x^2-x-1)的原函数是什么
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解：∵1/(x²-x-1)=1/[x-(1+√5)/2][x-(1-√5)/2]
=1/√5{1/[x-(1+√5)/2]-1/[x-(1-√5)/2]}
∴∫dx/(x²-x-1)=1/√5∫{1/[x-(1+√5)/2]-1/[x-(1-√5)/2]}dx
=1/√5{ln[x-(1+√5)/2]-ln[x-(1-√5)/2]}+C
=1/√5ln{[x-(1+√5)/2]/[x-(1-√5)/2]}+C
=1/√5ln{[2x-(1+√5)]/[2x-(1-√5)]}+C (C是积分常数)
=1/√5{1/[x-(1+√5)/2]-1/[x-(1-√5)/2]}
∴∫dx/(x²-x-1)=1/√5∫{1/[x-(1+√5)/2]-1/[x-(1-√5)/2]}dx
=1/√5{ln[x-(1+√5)/2]-ln[x-(1-√5)/2]}+C
=1/√5ln{[x-(1+√5)/2]/[x-(1-√5)/2]}+C
=1/√5ln{[2x-(1+√5)]/[2x-(1-√5)]}+C (C是积分常数)
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