Question

高数,定积分,设∫上限x下限0 f '(ln t)dt=ln(1+x),且f(0)=0,求f(x)

Answer 1

∫上限x下限0 f '(ln t) dt = ln(1+x) 变上限积分求导：

f '(lnx) = 1/(1+x)

=> f '(x) = 1/(1+ e^x)

=> f(x) = ∫ 1/(1+e^x) dx = ∫ e^(-x) / [ e^(-x) +1] dx

= - ln[ e^(-x) +1] + C

f(0)=0 => 0 = - ln2 + C => C=ln2

=> f(x) = ln2 - ln[ e^(-x) +1]