一直符合函数f(e^)=e^+^,求不定积分∫f(^)d^.^为x.

2025-01-03 15:52:06
推荐回答(1个)
回答1:

f(e^x) = e^x + x
f(e^lnx) = e^lnx + lnx
f(x) = x + lnx
∫ f(x) dx
= ∫ x dx + ∫ lnx dx
= x²/2 + xlnx - ∫ x * 1/x dx
= x²/2 + xlnx - x + C