Question

设当x≠0时，f(x)连续，求∫(xf'(x)-(1+x)f(x))/(x^2×e^x)dx？

Answer 1

简单计算一下即可，答案如图所示

Answer 2

∫(xf'(x)-(1+x)f(x))/(x^2×e^x)dx
= ∫(f(x)/x)' - f(x)/x] e^(-x) dx
= [f(x)/x]e^(-x) + c
思路：[e^f(x) g(x)] = e^f(x) [f'(x)g(x) + g'(x)]