一直符合函数f(e^)=e^+^,求不定积分∫f(^)d^. ^为x.请高人解答!谢谢

2025-01-03 15:41:38
推荐回答(2个)
回答1:

设e^t=x
∫f(x)dx
=∫f(e^t)d(e^t)
=∫(e^t+t)(e^t)dt
=∫e^2tdt+∫t(e^t)dt
=∫(1/2)*2*e^2tdt+∫t(e^t)dt+∫e^tdt-∫e^tdt
=∫(1/2)*2*e^2tdt+∫(t(e^t)+e^t)dt-∫e^tdt
=(1/2)*e^2t+t(e^t)-e^t+C
=(1/2)*x^2+xlnx-x+C

回答2:

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