Question

设f'(lnx)=x+1,f(0)=0,求f(x).

Answer 1

f'(lnx)=x+1
换元,y=lnx,x=e^y
f'(y)=e^y+1
改写,f'(x)=e^x+1
再求不定积分
f(x)
=∫f'(x)dx
=∫(e^x+1)dx
=∫e^xdx+∫dx
=e^x+x+C
又有,f(0)=0
那么,f(x)=e^x+x-1
有不懂欢迎追问