Question

f(x+1)的不定积分是xeˇ(x+1)+c,求函数f(x)

Answer 1

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