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  
$$\int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \approx 2.577632915067858 $$
 A: 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.
A: You have:
$$\frac{(1+x)^{1+x}}{x^x}=(x+1)\left(1+\frac{1}{x}\right)^{x}$$
For  $x \in (0,1)$:
$$1+\frac{1}{x} \geq 1+\frac{1}{1}=2$$
So:
$$(x+1)\left(1+\frac{1}{x}\right)^{x} \geq (x+1)2^x$$
And:
$$\int_{0}^{1}(x+1)\left(1+\frac{1}{x}\right)^{x} \; dx \geq \int_{0}^{1}(x+1)2^x \; dx$$
It's easy to calculate $\int_{0}^{1}(x+1)2^x \; dx$, it,s $\approx 2,24$.
Next using Bernoulli's inequality there is:
$$(1+\frac{1}{x})^{x} \leq 1+x \cdot \frac{1}{x}=2$$ so:
$$(x+1)\left(1+\frac{1}{x}\right)^{x} \leq 2(x+1)$$
And finally:
$$\int_{0}^{1}(x+1)\left(1+\frac{1}{x}\right)^{x} \; \leq \int_{0}^{1}2(x+1) \; dx =3$$
A: The integrand function is concave on $[0,1]$, hence the integral exceeds
$$\frac{f(0)+f(1)}{2}=\frac{5}{2}.$$
For the upper bound, it is sufficient to consider that $f'(1)=4\log 2$, hence the integral is less than
$$ \int_{0}^{1}\left(4+4\log 2(x-1)\right)\,dx = 4-2\log 2 <\frac{8}{3}.$$
A: For the left-hand inequality, you can use the integral form of Jensen's inequality. Since $e^x$ is convex, you can say
$$e^{\int_0^1 g(x)\,dx} \leq \int_0^1 e^{g(x)}\,dx.$$
where $g(x)$ is the natural logarithm of your integrand. You can compute $\int_0^1 g(x)\,dx$ explicitly to be $-1/2+\ln 4$. Then
$$e^{\int_0^1 g(x)\,dx} = e^{-1/2+\ln 4} = \frac{4}{\sqrt{e}}\geq\frac{4}{\sqrt{4}}=2.$$
