Question

高数，定积分，求解！！！

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]

Answer 2

f'(lnx)/x=1/(x+1)
=>f'(lnx)=x/(x+1)
令 lnx=t
f'(t)=e^t/(e^t+1)
=>f(t)=ln(e^t+1)+C
f(0)=0
=>f(0)=ln2+C=>C=-ln2
=>f(x)=ln(e^x+1)-ln2
希望对你有帮助

Answer 3

729707767的答案对的