<s>[INST] Prove $\int_{0}^{1}{{2x+\ln{1-x\over 1+x}\over x^3(1-x^2)^{1/2}}}dx=-(\pi/2)^2$ and $\int_{0}^{1}{2x+\ln{1-x\over 1+x}\over x^2(1-x^2)^{1/2}}dx=-2$ Motivated by this Question
Two similar integrals, but exhibit completely different closed forms

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

and 

$$\int_{0}^{1}{{2x+\ln\left({1-x\over 1+x}\right)\over x^2\sqrt{1-x^2}}}\mathrm dx=-\color{red}{2}\tag2$$

Here is my try:
I don't think substitution work here initially, but anyway I try
$u={1-x\over 1+x}\implies -{(1+x)^2\over 2}du=dx$, then $(1)$ becomes after a lengthy simplication to
$$\sqrt{2}\int_{0}^{1}{2\left({1-u\over 1+u}\right)+\ln u \over \sqrt{u}(1-u)^3}\cdot(1+u)^{11/2}\mathrm du\tag3$$
Or we split $(1)$ into
$$2\int_{0}^{1}{\mathrm dx\over x^2\sqrt{1-x^2}}+\int_{0}^{1}{\ln(1-x)\over x^3\sqrt{1-x^2}}\mathrm dx-\int_{0}^{1}{\ln(1+x)\over x^3\sqrt{1-x^2}}\mathrm dx=I_1+I_2-I_3\tag4$$
$I_1$ diverges.
How can we prove integrals $(1)$ and $(2)$?
 [/INST] Here's a possible path to #1 with some arguably important details left out.  So I'm making this community wiki in case someone else wants to come along and fill in those details or just completely fix this.
Recall:
$$ \ln(1+x) = \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}n x^n$$
So then:
$$ \ln(1-x) = \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}n (-x)^n = \sum_{n=1}^{+\infty} \frac{-1}n x^n$$
And:
\begin{align*}
  2x + \ln\left(\frac{1-x}{1+x}\right) &= 2x + \ln(1-x) - \ln(1+x)\\[0.3cm]
    &= 2x + \sum_{n=1}^{+\infty} \frac{-1}n x^n - \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}n x^n\\[0.3cm]
    &= 2x + \left(-x -\frac12x^2 -\frac13x^3 -\cdots\right) - \left(x - \frac12 x^2 + \frac13 x^3 - \cdots\right)\\[0.3cm]
    &= 2x  -2 \left(x + \frac13x^3 + \frac15x^5 + \cdots\right)\\[0.3cm]
    &= -2\sum_{n=1}^{+\infty} \frac{x^{2n+1}}{2n+1}
\end{align*}
Then (sweeping some convergence details under the rug here) we have:
\begin{align*}
\int_0^1 \frac{2x + \ln\left(\frac{1-x}{1+x}\right)}{x^3\sqrt{1-x^2}} \, dx &= \int_0^1 \frac{\displaystyle -2\sum_{n=1}^{+\infty} \frac{x^{2n+1}}{2n+1}}{x^3\sqrt{1-x^2}} \, dx\\[0.3cm]
  &= -2\sum_{n=1}^{+\infty} \frac1{2n+1} \int_0^1 \frac{x^{2n-2}}{\sqrt{1-x^2}} \, dx
\end{align*}
Let $x = \sin \theta$.  Then $$ \int_0^1 \frac{x^{2n-2}}{\sqrt{1-x^2}} \, dx = \int_0^{\pi/2} \sin^{2n-2} \theta \, d\theta. $$
Wolfram Alpha says: $$\int_0^{\pi/2} \sin^{2n-2} \theta \, d\theta = \frac{\sqrt\pi \ \Gamma\left(n-\frac12\right)}{2\Gamma(n)}$$
I suspect this can be proved by induction but I didn't quite get there.  Anyway, now we have:
$$
\int_0^1 \frac{2x + \ln\left(\frac{1-x}{1+x}\right)}{x^3\sqrt{1-x^2}} \, dx = -2\sum_{n=1}^{+\infty} \frac{\sqrt\pi \ \Gamma\left(n-\frac12\right)}{2(2n+1) \Gamma(n)}$$
Not quite sure how to evaluate that summation or if it's even possible.  But it does give the correct value.
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