设t=x+1,则x=t-1,代入∫f(x+1)dx = xeˇ(x+1)+c,得到
∫f(t)dt = (t-1)e^t+c
对上式求导得
f(t) = e^t + (t-1)e^t = te^t
即 f(x) = xe^x