d(f(1/x))/dx=2/x 求f(1/2)

d(f(1/x))/dx=2/x
两边积分得
f(1/x)=2lnx+C
不对啊,怎么有个常数?

d(f(1/x))/dx = f '(1/x) * (-1/x²) = 2/x
f '(1/x) = -2x
=> f '(x) = - 2/x
=> f (x) = - 2 ln|x| + C
f(1/2) = 2 ln2 + C

两边积分得，f(1/x)=2ln(1/x),则 f（/2）=-2ln2