设f(x)为可导函数,且满足f(x)=∫（上限X下线1）f(t)/tdt+(x-1)e^x求f(x)

f(x) = ∫(1,x) f(t)/t dt + (x - 1)e^xf'(x) = f(x)/x * dx/dx + e^x * (1 - 0) + (x - 1) * e^xf'(x) = f(x)/x + xe^x==> y' = y/x + xe^x==> y' - y/x = xe^x、e^(∫ - 1/x dx) = e^(- lnx) = 1/x、将1/x乘以方程...