设f'(x)在[0,1]上连续,试求∫[1+xf'(x)]e^f'(x)dx(范围是0到1)抱歉，输入有误，e^f(x)更正为e^f(x)，即原题为：设f'(x)在[0，1]上连续，试求∫[1+xf'(x)]e^f(x)dx(范围是0到1)

Question

设 f'(x)在[0,1]上连续,试求∫[1+x f'(x)]e^ f'(x)dx ( 范围是0到1)
抱歉，输入有误，e^ f(x)更正为e^ f(x)，即原题为：
设 f'(x)在[0，1]上连续，试求∫[1+x f'(x)]e^ f(x)dx ( 范围是0到1)

Answer 1

∫[1+x f'(x)]e^ f(x)dx
=∫e^ f(x)dx+∫xf'(x)e^ f(x)dx ‘先将括号打开,拆为两个积分
=∫e^ f(x)dx+∫xe^ f(x)df(x)
=∫e^ f(x)dx+∫xde^ f(x)
=∫e^ f(x)dx+xe^ f(x)-∫e^ f(x)dx '分部积分法则
=xe^ f(x)
最后结果1*e^ f(1)-0*e^ f(0)=e^ f(1)