<s>[INST] Prove that $2 \le \int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \le 3$ I need some starting ideas, hints for proving that 
$$2 \le \int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \le 3$$
I already checked that with Mathematica that numerically says that  
$$\int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \approx 2.577632915067858 $$
 [/INST] This can be done with 4 observations:
$$
\lim_{x\rightarrow 0^+} \frac{(1+x)^{1+x}}{x^x} = 1
$$
$\frac{(1+x)^{1+x}}{x^x}$ has value $4$ at $x=1$.
$\frac{(1+x)^{1+x}}{x^x}$ has value $\frac{\sqrt{27}}{2}$ at $x=\frac{1}{2}$.
At $x=1$, $\frac{d^2}{dx^2} \frac{(1+x)^{1+x}}{x^x} = 4 \ln 2$.
And $\frac{d^2}{dx^2} \frac{(1+x)^{1+x}}{x^x} < 0 $ on $(0,1)$, that is, the function is concave downward in the relevant interval.
Consider the trapezoid with corners at the $(0,0), (0,1), (1,4), (1,0)$ which has area $2.5$:  Because $f(x) = \frac{(1+x)^{1+x}}{x^x}$ is concave downward, the line between $(0,1)$ and $(1,4)$ always lies on or above $y=f(x)$.  This shows that the integral 
$$
I > 2.5
$$
Now consider the trapezoid with corners at the $(0,0), (0,4-4\ln 2), (1,4), (1,0)$ 
which has area $4 - 2 \ln 2$.   Because $f(x) = \frac{(1+x)^{1+x}}{x^x}$ is concave downward, the line between $(0,4-2\ln 2)$ and $(1,4)$ is always on or above $y=f(x)$.  This shows that the integral
$$
I < 4 - 2 \ln 2 
$$
Since $4 - 2 \ln 2 < 2.62$ this shows that the integral is between 2.50 and 2.62.
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