Showing $\int_{0}^{1}{1-x^2\over x^2}\ln\left({(1+x^2)^2\over 1-x^2}\right)dx=2$ Showing 

$$I=\int_{0}^{1}{1-x^2\over x^2}\ln\left({(1+x^2)^2\over 1-x^2}\right)dx=2\tag1$$

$$I=\int_{0}^{\infty}{1-x^2\over x^2}\ln\left({1+x^2\over 1-x^2}\right)+\int_{0}^{\infty}{1-x^2\over x^2}\ln(1+x^2)dx\tag2$$
Let $$J=\int_{0}^{\infty}{1-x^2\over x^2}\ln\left({1+x^2\over 1-x^2}\right)dx\tag3$$
Apply integration by part to (3)
$u=\ln\left({1+x^2\over 1-x^2}\right)\rightarrow du={4x\over (1-x^2)^2}dx$
$dv=1-x^{-2}\rightarrow v=x-x^{-1}$
$$J=\left.-{x+1\over x}\ln\left({1+x^2\over 1-x^2}\right)\right|_{0}^{1}+\int_{0}^{1}{4\over (1+x)(1-x)^2}dx\tag4$$
$$J=\left.-{x+1\over x}\ln\left({1+x^2\over 1-x^2}\right)\right|_{0}^{1}+\int_{0}^{1}\left({1\over 1+x}+{1\over 1-x}+{2\over (1-x)^2}\right)dx\tag5$$
$$J=\left.-{x+1\over x}\ln\left({1+x^2\over 1-x^2}\right)\right|_{0}^{1}+\left.\ln\left({1+x\over 1-x}\right)-{2\over 1-x}\right|_{0}^{1}\tag6$$
I am going to stop here. Evaluate these limits is not valid because of ${1\over 0}$. I have try substitution it looks more messier and complicated than this. I need some help, thank.
 A: We have $$I=\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(\frac{\left(1+x^{2}\right)^{2}}{1-x^{2}}\right)dx$$ $$=2\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(1+x^{2}\right)dx-\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(1-x^{2}\right)dx
 $$ Let us analyze the first integral. We have $$\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(1+x^{2}\right)dx=\int_{0}^{1}\frac{\log\left(1+x^{2}\right)}{x^{2}}dx-\int_{0}^{1}\log\left(1+x^{2}\right)dx
 $$ and integrating by parts we get $$\int_{0}^{1}\frac{\log\left(1+x^{2}\right)}{x^{2}}dx=-\log\left(2\right)+2\int_{0}^{1}\frac{1}{1+x^{2}}=-\log\left(2\right)+\frac{\pi}{2}.
 $$ For the other integral we can integrate by parts again $$\int_{0}^{1}\log\left(1+x^{2}\right)dx=\log\left(2\right)-2\int_{0}^{1}\frac{x^{2}}{x^{2}+1}dx
 $$ $$=\log\left(2\right)+\frac{\pi}{2}-2
  $$ hence $$\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(1+x^{2}\right)dx=2-2\log\left(2\right).
 $$ Fro the second integral we can do essentially the same argument $$\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(1-x^{2}\right)dx=2-4\log\left(2\right)
 $$ hence $$\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(\frac{\left(1+x^{2}\right)^{2}}{1-x^{2}}\right)=2.$$
A: The Taylor series approach is quite straightforward, too. We have:
$$ 2\log(1+x^2)-\log(1-x^2)=\sum_{n\geq 1}\frac{1-2(-1)^n}{n}x^{2n} \tag{1}$$
so by multiplying the RHS by $\frac{1}{x^2}-1$ we get:
$$ \frac{1-x^2}{x^2}\left(2\log(1+x^2)-\log(1-x^2)\right)=3+\sum_{n\geq 1}\frac{-1+(2+4n)(-1)^n}{n(n+1)}x^{2n} \tag{2}$$
and by integrating over $(0,1)$ the original integral turns into:
$$ 3-\sum_{n\geq 1}\frac{1}{n(n+1)(2n+1)}+2\sum_{n\geq 1}\frac{(-1)^n}{n(n+1)}=3-(3-4\log 2)+(2-4\log 2)\tag{3}$$
i.e. $\color{red}{\large 2}$, by partial fraction decomposition. We may also notice that:

$$ -\sum_{n\geq 1}\frac{1}{n(n+1)(2n+1)}+\sum_{n\geq 1}\frac{2(-1)^n}{n(n+1)} \\= -1+\sum_{n\geq 1}\left(\frac{1}{n(2n+1)}-\frac{1}{(n+1)(2n+1)}-\frac{1}{n(n+1)(2n+1)}\right)=-1+\sum_{n\geq 1}\color{red}{0}\tag{4}$$

and avoid partial fraction decomposition at all!
