How do we prove that $\int_0^1 \ln x\left({1\over \ln{x}}+{1\over 1-x}\right)^2\,dx=\gamma-1?$ How do we prove that:

$$\int_{0}^{1}\ln{x}\left({1\over \ln{x}}+{1\over 1-x}\right)^2\, dx =\color{blue}{\gamma-1}?\tag1$$

The only idea came to mind was this series
$$\sum_{n=1}^{\infty}{1\over 2^k(1+x^{-1/2^k})}={x\over 1-x}-{1\over \ln{x}}\tag2$$
Or expanded $(1)$
$$\int_0^1 \left({1\over \ln x} + {2 \over 1-x}+{\ln x \over (1-x)^2} \right)\,dx=\gamma-1\tag3$$
$$\int_0^1 {\ln x \over (1-x)^2}\,dx=\sum_{n=0}^\infty (1+n)\int_0^1 x^n\ln x \,dx = \sum_{n=0}^\infty (1+n)\cdot{-1\over (1+n)^2}\tag4$$
But $(4)$ diverges!
$\int {1\over 1-x} \, dx=-\ln(1-x)$
$\int_0^1 {2\over 1-x} \, dx$ also diverges
$\int{1\over \ln x} dx = \ln(\ln x )+\ln x +{\ln^2 x\over 2\cdot2!}+{\ln^3 x \over 3\cdot 3!}+\cdots$
$\int_0^1 {1\over \ln x} \, dx$ diverges too.
How do we go about integrating $(1)$?
Help needed, thanks!

Observe that $$ \int_{0}^{1}\left(\frac{1}{1-x}+\frac{\log\left(x\right)}{\left(1-x\right)^{2}}\right)dx\stackrel{x\rightarrow1-x}{=}\int_{0}^{1}\left(\frac{1}{x}+\frac{\log\left(1-x\right)}{x^{2}}\right)dx.$$ Fix $0<a<1$. We have $$I(a)=\int_{a}^{1}\left(\frac{1}{x}+\frac{\log\left(1-x\right)}{x^{2}}\right)dx
 $$ $$=\frac{-a\log\left(a\right)-\left(a-1\right)\log\left(1-a\right)+a\log\left(a\right)}{a}=-\frac{\left(a-1\right)\log\left(1-a\right)}{a}\underset{a\rightarrow0^{+}}{\rightarrow}-1.$$ The identity $$\int_{0}^{1}\left(\frac{1}{\log\left(x\right)}+\frac{1}{1-x}\right)dx=\gamma$$ is classical. See for example here.
