不好意思，后面的条件没有看懂
df/dx =f^2-6f+8
∫df/(f^2-6f+8) =∫dx
∫df/[(f-2)(f-4)] =x
(1/2)∫[1/(f-4) - 1/(f-2) ]df =x
(1/2)ln| (f-4)/(f-2)| = x+C
f(x0)=2
(1/2)ln| (x0-4)/(x0-2)| = 2+C
C = (1/2)ln| (x0-4)/(x0-2)| -2
(1/2)ln| (f-4)/(f-2)| = x+(1/2)ln| (x0-4)/(x0-2)| -5
x = (1/2)ln| (f-4)/(f-2)| +2 -(1/2)ln| (x0-4)/(x0-2)|
f(x0)=3
x = (1/2)ln| (f-4)/(f-2)| +3 -(1/2)ln| (x0-4)/(x0-2)|
f(x0)=4
x = (1/2)ln| (f-4)/(f-2)| +4 -(1/2)ln| (x0-4)/(x0-2)|
f(x0)=5
x = (1/2)ln| (f-4)/(f-2)| +5 -(1/2)ln| (x0-4)/(x0-2)|
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