作业帮f(x)=x^2+∫[0~x]e^（x-t)f '(t)dt 怎么变到 f '(x)=2x+f '(x)+∫[0~x]e^（x-t)f '(t)dt
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![f(x)=x^2+∫[0~x]e^（x-t)f '(t)dt 怎么变到 f '(x)=2x+f '(x)+∫[0~x]e^（x-t)f '(t)dt](/uploads/image/z/12512715-51-5.jpg?t=f%28x%29%3Dx%5E2%2B%E2%88%AB%5B0%7Ex%5De%5E%EF%BC%88x-t%29f+%27%28t%29dt+%E6%80%8E%E4%B9%88%E5%8F%98%E5%88%B0+f+%27%28x%29%3D2x%2Bf+%27%28x%29%2B%E2%88%AB%5B0%7Ex%5De%5E%EF%BC%88x-t%29f+%27%28t%29dt)
f(x)=x^2+∫[0~x]e^（x-t)f '(t)dt 怎么变到 f '(x)=2x+f '(x)+∫[0~x]e^（x-t)f '(t)dt
f(x)=x^2+∫[0~x]e^（x-t)f '(t)dt 怎么变到 f '(x)=2x+f '(x)+∫[0~x]e^（x-t)f '(t)dt
f(x)=x^2+∫[0~x]e^（x-t)f '(t)dt 怎么变到 f '(x)=2x+f '(x)+∫[0~x]e^（x-t)f '(t)dt
f(x)=x²+∫[0→x] e^(x-t) f '(t) dt
=x²+e^x∫[0→x] e^(-t) f '(t) dt
两边同时对x求导：
f '(x)=2x+e^x∫[0→x] e^(-t) f '(t) dt+e^x*e^(-x) f '(x)
=2x+e^x∫[0→x] e^(-t) f '(t) dt+f '(x)
=2x+∫[0→x] e^(x-t) f '(t) dt+f '(x)