Question

这道积分题？

求大佬给个思路...自己算的有点迷.

Answer 1

令 √(2e^y-3e) = u, 则 e^y = (u^2+3e)/2, y = ln(u^2+3e)-ln2
dy = 2udu/(u^2+3e), 得
I = 2∫[1+ln2-ln(u^2+3e)]du/(u^2+3e)
= 2(1+ln2)∫du/(u^2+3e) - 2∫ln(u^2+3e)]du/(u^2+3e)
= [2(1+ln2)/√(3e)]arctan[u//√(3e)] - 2∫ln(u^2+3e)]du/(u^2+3e)
令 u = √(3e)tanv, 则 du = √(3e)(secv)^2dv
I1 = ∫ln(u^2+3e)]du/(u^2+3e)
= ∫ln[3e(secv)^2]√(3e)(secv)^2dv/[3e(secv)^2]
= [1/√(3e)]∫ln[3e(secv)^2]dv = [1/√(3e)]∫ln(3e)-2ln(cosv)]dv
= vln(3e)/√(3e) - [2/√(3e)]∫ln(cosv)dv
= vln(3e)/√(3e) - [2/√(3e)]vln(cosv) + [2/√(3e)]∫[v(-sinv)/(cosv)]dv
= vln(3e)/√(3e) - [2/√(3e)]vln(cosv) - [2/√(3e)]∫vtanvdv
好像不能能用初等函数表示 ？

Answer 2

将分母先化简。
分子是 1-y? 如果是1+y就很好做了，可用换元法d(yeʸ)进行换元。

追问

有道理.但他偏偏就是1-y...