Is it always a positive value? The values of $x$ can be from $0$ to $\infty$. So for these values of $x$ is the following function always positive $$f(x)=x^2+x+\exp(\frac{1}{x})(2x+1)Ei(-\frac{1}{x})$$ where $Ei(x)$ is the exponential integral. My checking in the WA shows that it is a positive value for arbitrarily large values of $x$.
 A: Nobody likes the exponential integral, but luckily, it is linked to the realm of well-known things by a bridge called derivative. So can't we turn to using its derivative instead?
To begin with, let's multiply $f$ by a certain positive function and thus switch to an equivalent question: is $g$ always positive for $x>0$, where 
$$g(x)={x^2+x\over2x+1}e^{-1/x}+Ei\left(-{1\over x}\right)=\left({x\over2}+{1\over4}-{1\over4(2x+1)}\right)e^{-1/x}+Ei\left(-{1\over x}\right)$$
Well, $g$ is undefined at $0$ (as is your $f$), so both have to be defined there by continuity, and it is quite obvious that $\lim\limits_{x\to0}g(x)=0$.
Now comes the derivative.
$$g'(x)=
\left({1\over2}+0+{1\over2(2x+1)^2}\right)e^{-1/x}
+{x^2+x\over2x+1}\cdot{1\over x^2}e^{-1/x}
-{e^{-1/x}\over x}=\\=
e^{-1/x}\cdot{1\over(2x+1)^2}\cdot\left({(2x+1)^2\over2}+{1\over2}+2x+1+{2x+1\over x}-{(2x+1)^2\over x}\right)=\\=
e^{-1/x}\cdot{1\over(2x+1)^2}\cdot\left(2x^2+2x+{1\over2}+{1\over2}+2x+1-4x-2\right)=\\=
e^{-1/x}\cdot{2x^2\over(2x+1)^2}\color{red}{\Large\bf>0}
$$
Hence $g(x)>0$ for $x>0$, and so is $f(x)$.
Q.e.d.
A: \begin{align}
\mathrm{Ei}(-1/x) &= - \int_{1/x}^\infty \frac{e^{-t}}{t} \, dt \\
f(x) &> 0 \\
x (x + 1) &> - (2x+1) \exp(1/x) \, \mathrm{Ei}(−1/x) \\
x (x + 1) &> (2x+1) \exp(1/x) \, \int_{1/x}^\infty \frac{e^{-t}}{t} \, dt \\
\end{align}
Since by requirement $x$ is positive, we can bring everything except the integral to the left without changing the direction of the inequality sign
\begin{align}
\frac{x(x + 1)}{(2x + 1)} \exp(-1/x) &> \int_{1/x}^\infty \frac{e^{-t}}{t} \, dt \\
\end{align}
now write everything in terms of $u = 1/x$:
\begin{align}
\frac{(1 + u)}{u(2 + u)} \exp(-u) &> \int_{u}^\infty \frac{e^{-t}}{t} \, dt \\
\end{align}
take the difference of both sides $\Delta(u)$ and take the derivative with respect to $u$
\begin{align}
\Delta'(u) &= \left[ \frac{u(2 + u) - 2(1+u)^2}{u^2(2 + u)^2}- \frac{(1 + u)}{u(2 + u)} \right] \exp(-u) + \frac{e^{-u}}{u}  \\
&= \left[ - \frac{(u^2 + 2u + 2) + u(1+u)(2+u)}{u^2(2 + u)^2} \right] \exp(-u) + \frac{e^{-u}}{u}  \\
&= \left[ 1 - \frac{2 + 4u + 4u^2 + u^3}{u(2 + u)^2} \right] \frac{e^{-u}}{u}  \\
&= \left[ \frac{(2u + 4u^2 + u^3) -  (2 + 4u + 4u^2 + u^3)}{u(2 + u)^2} \right] \frac{e^{-u}}{u}  \\
&= - \frac{ 2 (1 + u) }{u^2(2 + u)^2} e^{-u}  \\
\end{align}
since $x > 0$ we have $u > 0$, so this difference function is strictly monotonically decreasing. The divergence approaches $\Delta'(u) \to 0$ very quickly for $u \to \infty$, so $\Delta(u)$ converges to a minimum at $u \to \infty$ or $x \to 0$.
So all we need to check is whether $f(x)$ is non-negative for $x \to 0$. Clearly we can ignore the polynomial terms, so we get 
\begin{align}
f(0) &= \lim_{x\to0} \, e^{1/x} \int_{1/x}^\infty \frac{e^{-t}}{t} \, dt
\end{align}
which is clearly non-negative. Thus $f(x)$ is positive for all $x > 0$.
