f(x)在[1,+∞）内有连续的导数,且满足x-1+x∫(上限x,下限1)f(t)dt=(x+1)∫(上限x,下限1)tf(t)dt,求f(x)

原方程可化为：x-1 = x*( [1,x]∫t*f(t)dt) + [1,x]∫t*f(t)dt - x*( [1,x]∫f(t)dt ) =x*( [1,x]∫(t-1)*f(t)dt ) + [1,x]∫t*f(t)dt --- (1)设 F1(t) = ∫(t-1)*f(t)dt ,F2(t) = ∫t*f(t)dt,则：F1'(t) = (t-1)*f(...