<s>[INST] Squeeze an integral Would you have any idea about this problem ?

Prove that for all nonnegative integers $n$, the following inequalities hold:
  $$\frac{e^2}{n+3}\leq \int_{1}^{e} x (\ln x)^n \,dx \leq \frac{e^2}{n+2} ,$$ 
  where $\ln$ is the natural logarithm.

Thank you.
 [/INST] The rightmost inequality can be proved using substitution and integration by parts.
In fact, set $I(n):=\int_1^e x\ (\ln x)^n\ \text{d} x$; then:
$$I(n) \stackrel{t=\ln x}{=} \int_0^1 e^{2t}\ t^n\ \text{d} t$$
and integrating by parts three times the RH side yields:
$$\tag{1} I(n)= \frac{e^2 (n^2+3n+4)}{(n+1)(n+2)(n+3)} - \frac{8}{(n+1)(n+2)(n+3)}\ I(n+3)\; ;$$
since $I(\cdot) \geq 0$ and $n^2+3n+4\leq (n+1)(n+3)$, you get:
$$I(n)\leq \frac{e^2 (n^2+3n+4)}{(n+1)(n+2)(n+3)} \leq \frac{e^2}{n+2}\; .$$
I didn't make the computations, but I bet your leftmost inequality can be proved in analogous manner... It seems all you have to do is another integration by parts in (1).

Following Didier advice, I complete the answer.
Once you got the the upper bound for $I(n)$, you have two ways to obtain the desidered lower bound for $I(n)$.


*

*Didier's way: use the upper bound for $I(n+3)$ to lower bound $I(n)$. 
Since $I(n+3)\leq \frac{e^2}{n+5}$, from (1) and $8/(n+5)\leq 2$ you gain:
$$\begin{split}I(n) &\geq \frac{e^2 (n^2+3n+4)}{(n+1)(n+2)(n+3)} - \frac{8}{(n+1)(n+2)(n+3)}\ \frac{e^2}{n+5}\\ &\geq \frac{e^2(n^2+3n+4)}{(n+1)(n+2)(n+3)} -\frac{2e^2}{(n+1)(n+2)(n+3)} \\ &= \frac{e^2 (n^2+3n+2)}{(n+1)(n+2)(n+3)}\\ &\geq \frac{e^2}{n+3}\; .\end{split}$$

*Pacciu's way: use again integration by parts. 
In fact, another integration by parts in (1) yields:
$$I(n)=\frac{e^2(n^3+7n^2+16n+8)}{(n+1)(n+2)(n+3)(n+4)}+\frac{16}{(n+1)\cdots (n+4)}\ I(n+4)$$
therefore you can use $I(\cdot )\geq 0$ and $n^3+7n^2+16n+8 \geq (n+1)(n+2)(n+4)$ to get your lower bound.
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