∫(0,x)f`(lnt)dt=ln(1+x)且f(0)=0,求f(x).

∫(0,x)f`(lnt)dt=ln(1+x)且f(0)=0,求f(x).
∫(0,x)f`(lnt)dt=ln(1+x)且f(0)=0,求f(x).

∫(0,x)f`(lnt)dt=ln(1+x)且f(0)=0,求f(x).
∫(0~x) f'(lnt) dt = ln(1 + x),f(0) = 0
两边求导
f'(lnx) = 1/(1 + x),令x = e^x
f'(x) = 1/(1 + e^x),之后两边求不定积分
f(x) = ∫ dx/(1 + e^x) = ∫ (1 + e^x - e^x)/(1 + e^x) dx = ∫ dx - ∫ d(1 + e^x)/(1 + e^x)
f(x) = x - ln(1 + e^x) + C
f(0) = 0 => 0 = - ln(2) + C => C = ln(2)
∴f(x) = x - ln(1 + e^x) + ln(2)