Extremely tough indefinite integral This integral does indeed use special functions, so do include them here. Evaluate: 

$\int \frac{1}{\sqrt{x}\ln(x)} dx$

$x = {\sqrt{x}}^{2} \space \text{let} \space u = \sqrt{x}$

$= 2\int \frac{1}{\ln(u^2)} du$

Now, I am not very experienced with special functions such as $Ei, li, Li$, etc...

Help please?
Hint

Change variable $x=e^{2t}$ and arrive to $$I=\int \frac{1}{\sqrt{x}\ln(x)} dx=\int \frac{e^t}{t} dt=\text{Ei}\left(t\right)$$

Puiseux expansion along the positive real axis is given by $$\text{Ei}\left(t\right)=\gamma+\log(|t|)+\sum_{k=1}^{\infty}\frac{t^k}{k~~ k!}$$

For example, for $t=e^2$, the expansion gives an error of $0.1$% using $13$ terms and $0.01$% after $18$ terms.
You're looking for a primitive I guess:

Let F(t) = $= \int \frac{u^t}{\ln(u)} du$

Prove that you can differentiate under the integral sign, and you get:

F'(t) = $= \int u^t du$ , F' can be easily calculated. You can then integrate on the "t" parameter and take the value of F in 0. I did not do the calculation all the way but it should work .