$I(a):= \int _1^e \frac{Lnx}{(a+Lnx)^{a+1}} dx$
اول فرض کنید $a \neq -1$:
$u:=a+Lnx \Rightarrow du= \frac{dx}{x} \Rightarrow dx=xdu,x=e^{u-a} ,u(1)=a,u(e)=a+1$
$ \Rightarrow I= \int _a^{a+1} \frac{(u-a)(e^{u-a})}{u^{a+1}}du= \frac{1}{e^a} [\int _a^{a+1} \frac{e^u}{u^a}du-a\int _a^{a+1} \frac{e^u}{u^{a+1}}du]$
$= \frac{1}{e^a}[ \int _a^{a+1}u^{-a}d(e^u)-a \int _a^{a+1}u^{-a-1}e^udu] $
$= \frac{1}{e^a} [u^{-a}e^u|_a^{a+1}+a\int _a^{a+1}u^{-a-1}e^udu-a\int _a^{a+1}u^{-a-1}e^udu] = \frac{1}{e^a} [u^{-a}e^u|_a^{a+1}]$
$= \frac{e}{(a+1)^a} - \frac{1}{a^a} $
اگر $a=-1$ داریم:
$I(-1)= \int _1^eLnxdx=[xLnx-x]_1^e=eLne-e-1.Ln1+1=1$
$ \Box $
