Inequality for the integral $\frac{\ln x}{x^n}$ Define the integral $I_{n}$ as follows for $n$ an integer greater than $1$:
$I_{n}:=\int_{1}^{e}\frac{\ln x}{x^n}dx$
Is it true that 

$$I_{n}\leq \frac{1}{n-1}\left(1-\frac{1}{e^{n-1}}\right)?$$

 A: Hint: See Davide Giraudo's answer to your last question.

For $n>1$:

$$
\int_1^e{1\over x^n}\,dx= {x^{-n+1}\over -n+1}\biggl|_1^e={e^{-n+1}\over -n+1} -{1\over -n+1}={1\over n-1}(1-e^{-n+1}).
$$
Now use $ 0\le\ln x\le 1$ for $1\le x\le e$ to obtain
$$
\int_1^e{\ln x\over x^n}\,dx\le \int_1^e{1\over x^n}\,dx={1\over n-1}(1-e^{-n+1}).
$$

Or, you can evaluate the integral exactly:
$$\eqalign{
\int_1^e \underbrace{(\ln x)\vphantom{x^n\over x^n1}}_u \,\,\underbrace{x^{-n}\,dx\vphantom{1\over x}}_{dv}
&= \underbrace{\ln x\vphantom{x^n\over1 x^n}}_u\,\,\underbrace{ {x^{-n+1}\over -n+1}}_{v}\biggl|_1^e - \int_1^e \underbrace{{x^{-n+1}\over -n+1}}_v\,\,\underbrace{ {1\over x}\,dx}_{du}\cr
&={ e^{-n+1}\over -n+1} +{1\over n-1}\int_1^e x^{-n}\,dx\cr
&={ e^{-n+1}\over -n+1} +{1\over n-1}\cdot {1\over n-1}(1-e^{-n+1})\cr
&={1\over n-1}\Bigl( -e^{-n+1} +{1\over n-1}(1-e^{-n+1})\Bigr)\cr
&={1\over n-1}\Bigl( {1\over n-1} -{n\over n-1} e^{-n+1}\Bigr)\cr
(&\le{1\over n-1}\Bigl( 1 -  e^{-n+1}\Bigr).)\cr
}
$$
