$\displaystyle t=\sqrt[x]{2\sqrt[x]{2\sqrt[x]{2...}}}$
$\displaystyle t=\sqrt[x]{2t}\Rightarrow t^x = 2t \Rightarrow t = 2^{\frac{1}{1-x}}$
$\displaystyle 2^{\frac{1}{1-x}} = e^{ln(2^{\frac{1}{1-x}})}= e^{\frac{1}{1-x}ln(2)}$
Amer's advice in post 4 gives you $\displaystyle \displaystyle \sqrt[x]{2\sqrt[x]{2\sqrt[x]{2\sqrt[x]{\dots}}}} = e^{\frac{\ln{2}}{1 - x}}$, so
$\displaystyle \displaystyle \begin{align*} \int{\frac{\sqrt[x]{2\sqrt[x]{2\sqrt[x]{2\sqrt[x]{\dots}}}}}{(1 - x)^2}\,dx} &= \int{\frac{e^{\frac{\ln{2}}{1 - x}}}{(1 - x)^2}\,dx} \\ &= \frac{1}{\ln{2}}\int{e^{\frac{\ln{2}}{1 - x}}\left[\frac{\ln{2}}{(1 - x)^2}\right]\,dx} \end{align*}$
Which now can be solved using the substitution $\displaystyle \displaystyle u = \frac{\ln{2}}{1 - x}$.
In future I suggest you read and try to understand and use the help that you are given instead of telling these people that their advice is "not relevant" and demanding an answer.
The 'core' of the question is the computation of...
$\displaystyle f(x)= \lim_{n \rightarrow \infty} a_{n} $ (1)
... where $\displaystyle a_{n}$ satisfies the difference equation...
$\displaystyle a_{n+1}= (2\ a_{n})^{\frac{1}{x}}$ (2)
The limit (1), if a finite limit exists, is the solution of the equation...
$\displaystyle \Delta_{n}= a_{n+1}-a_{n}= f(a_{n})=0$ (3)
... where...
$\displaystyle f(\xi)= (2\ \xi)^{\frac{1}{x}}-\xi$ (4)
With a little of patience You find that the solution of (3) is...
$\displaystyle \xi= 2^{- \frac{1}{1-x}}$ (5)
... so that is...
$\displaystyle \int \frac{2^{- \frac{1}{1-x}}}{(1-x)^{2}}\ dx = \frac{1}{\ln 2} 2^{-\frac{1}{1-x}} + c$ (6)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$