Prove $\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^1\frac{\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}dx$ How to prove 

$$\int_0^1\frac{\ln(1+a^2x)}{1+a^2x^2}dx=\int_0^1\frac{\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}dx\tag{1}$$

Here is how I came up with this relation:
In this solution @Kemono Chen elegantly proved

$$\int_0^a\frac{\ln(1+ax)}{1+x^2}dx=\int_0^1\frac{a\ln(1+a^2x)}{1+a^2x^2}dx=\frac12\arctan a\ln(1+a^2)\tag{2}$$

and while trying to prove the identity in (2) starting from RHS, I ended up with the relation in (1). So any straightforward method to prove (1)? Plus any good applications of (1)?
The transformation of the integral in (2) was done by @Jack D'Aurizio here.

I will post my proof in the answer section soon and I am tagging "harmonic number" as the proof involves it in case you are curious. Thanks

UPDATE: If we let $\frac{1-x}{x}\mapsto x$ in (1) then combine with (2) we have 

$$\int_0^1\frac{1+a^2x}{1+a^2x^2}dx=\int_0^\infty\frac{\ln x}{a^2+(1+x)^2}=\frac1{2a}\arctan a\ln(1+a^2)\tag{3}$$


In other words we want to show that:
$$\color{blue}{\int_0^1 \frac{\ln\left(\frac{1-x}{1+a^2 x}\right)}{1+a^2 x^2}dx}=\color{red}{\int_0^1 \frac{\ln x}{1+a^2 x^2}dx}$$
This can be seen via the substitution:
$$\frac{1-x}{1+a^2 x}=t\Rightarrow x=\frac{1-t}{1+a^2 t}\Rightarrow dx=-\frac{1+a^2}{(1+a^2t)^2}dt$$
$$\Rightarrow \color{blue}{\int_0^1 \frac{\ln\left(\frac{1-x}{1+a^2 x}\right)}{1+a^2 x^2}dx}=\int_0^1 \frac{\ln t}{1+a^2 \frac{(1-t)^2}{(1+a^2t)^2}}\frac{1+a^2}{(1+a^2t)^2}dt\overset{t=x}=\color{red}{\int_0^1 \frac{\ln x}{1+a^2 x^2}dx}$$
