Integrate $\int_0^\infty\frac{1}{x}\ln\left(\left(\frac{1+x}{1-x}\right)^2\right)dx$ using known sum

As people pointed out in the comment section, the radius of convergence for this power series is $1$ so you should split the integral from $0$ to $1$ and $1$ to $\infty.$

$$\int_0^\infty\frac{1}{x}\ln\left(\left(\frac{1+x}{1-x}\right)^2\right)dx=\int_0^1\frac{1}{x}\ln\left(\left(\frac{1+x}{1-x}\right)^2\right)dx+\int_1^\infty\frac{1}{x}\ln\left(\left(\frac{1+x}{1-x}\right)^2\right)dx \\ =\int_0^1\frac{1}{x}4\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)}dx+\int_1^\infty\frac{1}{x}\ln\left(\left(\frac{1+x}{1-x}\right)^2\right)dx \\ =4\sum_{k=0}^\infty\frac{1}{(2k+1)^2}+\int_1^\infty\frac{1}{x}\ln\left(\left(\frac{1+x}{1-x}\right)^2\right)dx=\frac{\pi^2}{2}+\int_1^\infty\frac{1}{x}\ln\left(\left(\frac{1+x}{1-x}\right)^2\right)dx$$

For the right side let $x=\frac{1}{t}$ then

$$\int_1^\infty\frac{1}{x}\ln\left(\left(\frac{1+x}{1-x}\right)^2\right)dx=\int_0^1\frac{1}{t}\ln\left(\left(\frac{1+t}{1-t}\right)^2\right)dt=4\sum_{k=0}^\infty\frac{1}{(2k+1)^2}=\frac{\pi^2}{2}.$$

So you get $$\int_0^\infty\frac{1}{x}\ln\left(\left(\frac{1+x}{1-x}\right)^2\right)dx=\frac{\pi^2}{2}+\frac{\pi^2}{2}=\pi^2.$$