How can we tackle this integral $\int_{0}^{1}{2x^2-2x+\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=-1?$ Something is wrong with this integral (in terms of splitting  them out)

$$\int_{0}^{1}{2x^2-2x+\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=\color{blue}{-1}\tag1$$

My try:
Splitting the integral
$$\int_{0}^{1}{2x^2-2x\over x^3\sqrt{1-x^2}}\mathrm dx+\int_{0}^{1}{\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=I_1+I_2\tag2$$
Note that $I_1$ and $I_2$ diverge, so how can we tackle it as a whole?
 A: I will present an answer without Gamma functions and without series' expansions. Instead, a more general problem is solved, where the OP's question is a special case.
Let 
$$
I(a) = \int_{0}^{1}{2a^2x^2-2ax+\ln[(1-ax)(1+ax)^3]\over x^3\sqrt{1-x^2}}\mathrm dx
$$
The answer to the OP's question is then given by $I(a = 1)$.
Then, after partial differentiation w.r.t. $a$, 
$$
I'(a) = \int_{0}^{1}   \frac{2a^2(2ax - 1)}{(a^2 x^2 - 1)\sqrt{1-x^2}}\mathrm dx
$$
Note that this removes the problematic factor $x^3$ in the denominator - this issue was solved in the previous solution by series' expansion of the $\log$. Hence, convergence is established and the order of integrations $x,a$ can be exchanged.
The $x$-integration gives
$$
I'(a) = a^2 \Big[ \frac{2 \arctan(\frac{x \sqrt{1 - a^2}}{\sqrt{1 - x^2}})}{\sqrt{1 - a^2}} + \frac{4\arctan(\frac{a \sqrt{1 - x^2}}{\sqrt{1 - a^2}})}{\sqrt{1 - a^2}} \Big]_{x=0}^{1}
$$
which is 
$$
I'(a) = a^2 \Big[ \frac{\pi}{\sqrt{1 - a^2}} -  \frac{4\arctan(\frac{a }{\sqrt{1 - a^2}})}{\sqrt{1 - a^2}} \Big] = a^2 \Big[ \frac{\pi}{\sqrt{1 - a^2}} -  \frac{4\arcsin({a })}{\sqrt{1 - a^2}} \Big]
$$
Now we integrate w.r.t. $a$:
$$
I(a) = \int I'(a) \rm{d} a  = \int a^2 \Big[ \frac{\pi}{\sqrt{1 - a^2}} -  \frac{4\arcsin({a })}{\sqrt{1 - a^2}} \Big] \rm{d} a
$$
Replacing $a = \sin (y)$ gives 
$$
I(y) = \int \sin^2(y) \Big[ {\pi} - {4 y}\Big] \rm{d} y = y \sin(2y) + (\pi y)/2 - \sin^2(y) - y^2 - (\pi \sin(2y))/4 + C
$$
We need the constant $C$. Obviously $0 = I(a=0) = I(y=0)$ which gives $C=0$. 
The function in question by the OP is $I(a=1) = I(y=\pi/2) = -1$.
This  solves the OP'S question. $\qquad \qquad \Box$ 
A: You have to split as
$$
I=\int_0^1\frac{x^2/2+x+\log(1-x)}{x^3\sqrt{1-x^2}}+\int_0^1\frac{3x^2/2-3x+3\log(1+x)}{x^3\sqrt{1-x^2}}=\color{red}{I_1}+\color{blue}{I_2}
$$
Now we can calculate
$$
\color{red}{I_1}=-\sum_{n=3}^{\infty}\frac{1}{n}\int_0^1\frac{x^{n-3}}{\sqrt{1-x^2}}=-\frac{\sqrt{\pi}}{2}\sum_{n=3}^{\infty}\frac{1}{n}\frac{\Gamma\left(\frac n2-1\right)}{\Gamma\left(\frac n2-\frac12\right)}\underbrace{=}_{(1)}\\
-\frac{1}{16}\sum_{n=3}^{\infty}\frac{2^n}{n}\frac{\Gamma^2\left(\frac n2-1\right)}{\Gamma\left(n-1\right)}=-\frac12\int_0^1dtt\sum_{k=1}^{\infty}(2t)^k\frac{ \Gamma^2 \left(\frac k2\right)}{k!}\underbrace{=}_{(2)}\\
\int_0^1dtt\arcsin(t)(\pi+\arcsin(t))=\color{red}{-\frac{1}{4}+\frac{3\pi^2}{16}}
$$
where we have used Legendre's duplication formula in $(1)$ and the results from this nice question in $(2)$ (the last integration is trivial after employing $t=\sin(q)$ so i spare it here).
Playing exactly the same game with $I_2$ we obtain
$$
\color{blue}{I_2}=-3\int_0^1dtt\arcsin(t)(\pi-\arcsin(t))=\color{blue}{-\frac{3}{4}-\frac{3\pi^2}{16}}
$$
or

$$
I=\color{red}{I_1}+\color{blue}{I_2}=-1
$$

