设f(x+x-1)=x3+x-3 g(x+x-1)=x2+x-2,求f(g(x))

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设f(x+x-1)=x3+x-3 g(x+x-1)=x2+x-2,求f(g(x))

设f(x+x-1)=x3+x-3 g(x+x-1)=x2+x-2,求f(g(x))
设f(x+x-1)=x3+x-3 g(x+x-1)=x2+x-2,求f(g(x))

设f(x+x-1)=x3+x-3 g(x+x-1)=x2+x-2,求f(g(x))
f(x+1/x)=x³+1/x³
g(x+1/x)=x²+1/x²
a=x+1/x
则a²=x²+2+1/x²
x²+1/x²=a²-2
所以g(a)=a²-2
x³+1/x³
=(x+1/x)(x²-1+1/x²)
=a(a²-1)
=a³-a
所以f(a)=a³-a
所以f[g(x)]
=(x²-2)³-(x²-2)
=x^6-6x^4+11x³-6

1/x(4-x)
=-(1/4)*4/(x-4)x
=-(1/4)*[x-(x-4)]/(x-4)x
=-(1/4)*[x/(x-4)x-(x-4)/(x-4)x]
=-(1/4)*[1/(x-4)-1/x]
所以原式=(-1/4)∫[1/(x-4)-1/x]dx
=(-1/4)(ln|x-4|-ln|x|)+C

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