Proving $\int_{0}^{e}\sqrt{\frac{1}{\operatorname{W}(x)}}-2\sqrt{\operatorname{W}(x)}dx=0$ for Lambert's $W$ function 
Prove that :
  $$\int_{0}^{e}\sqrt{\frac{1}{\operatorname{W}(x)}}-2\sqrt{\operatorname{W}(x)}dx=0$$
  where $\operatorname{W}(x)$ denotes the Lambert's $W$ function 

I can evaluate the integrals using antiderivatives:
$$\int_{0}^{e}\sqrt{\frac{1}{\operatorname{W}(x)}}dx=2\int_{0}^{e}\sqrt{\operatorname{W}(x)}dx=\frac{1}{2}\sqrt{\pi}\operatorname{(erfi}(1))+e$$
(where $\operatorname{erfi}(x)$ denotes the imaginary error function), but I would like to know if there is a tricky way or a better method to solve this. To start I think we can make the substitution $\operatorname{W}(x)=t$.
My question :

Have you an idea to solve this?

Thanks in advance! 
 A: Personally, I like this trick the most - it utilizes symmetry and can be proven by picture. Draw a function from the points $(a,f(a))$ to $(b,f(b))$. If the function is invertible and increasing we can say that
$$\int_a^b f(x)\:dx + \int_{f(a)}^{f(b)}f^{-1}(y)\:dy = b\cdot f(b) - a \cdot f(a)$$
in other words, the area between the curve and the $x$-axis plus the area between the curve and the $y$-axis equals the difference in area between those two rectangles. 
For the moment consider the integral
$$\int_0^e \sqrt{W(x)}\:dx$$
Inverting we get that
$$y = \sqrt{W(x)} \implies x = y^2e^{y^2}$$
And given that $W(0) = 0$, $W(e) = 1$ we have that
$$\int_0^e \sqrt{W(x)}\:dx = e - 0 - \int_0^1 y^2 e^{y^2}\:dy$$ $$ = e - \frac{1}{2}ye^{y^2}\Biggr|_0^1 + \frac{1}{2}\int_0^1e^{y^2}\:dy \equiv \frac{2e+\sqrt{\pi}\operatorname{erfi}(1)}{4}$$
A similar trick will work for the second integral, except the function is now decreasing. Drawing another picture, you will see that they are accumulating the same area, except one of them has an extra box.
A: Making it short after @Ninad Munshi'answer.
As @Ninad Munshi did, I prefered to use $\sqrt{W(x)}=t$ which, at the end, makes the antiderivative to be
$$-\frac{2 x (W(x)-1)}{\sqrt{W(x)}}$$
Edit
If I am not mistaken,
$$\int W^a(x)\,dx=\int e^t (t+1) t^a\,dt=(-1)^{-a} (\Gamma (a+1,-t)-\Gamma (a+2,-t))$$
$$\int_0^e W^a(x)\,dx=\int_0^1 e^t (t+1) t^a\,dt=$$ $$(-1)^{-a} (\Gamma (a+1,-1)-\Gamma (a+1,0)-\Gamma (a+2,-1)+\Gamma (a+2,0))$$
