<s>[INST] 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 

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

 [/INST] In the post body we have 
$$\int_0^1\frac{a\ln(1+a^2x)}{1+a^2x^2}dx=\frac12\arctan a\ln(1+a^2)\tag{*}$$
and from this solution we have 
\begin{align}
f(a)&=\frac12\arctan a\ln(1+a^2)=-\sum_{n=0}^\infty \frac{(-1)^n H_{2n}}{2n+1}a^{2n+1}\\
&=-\sum_{n=0}^\infty \frac{(-1)^n H_{2n+1}}{2n+1}a^{2n+1}+\sum_{n=0}^\infty \frac{(-1)^n }{(2n+1)^2}a^{(2n+1)}\\
&=\int_0^1a\ln(1-x)\sum_{n=0}^\infty(-a^2x^2)^n-\int_0^1a\ln x\sum_{n=0}^\infty(-a^2x^2)^n\\
&=\int_0^1\frac{a\ln(1-x)}{1+a^2x^2}\ dx-\int_0^1\frac{a\ln x}{1+a^2x^2}\ dx\\
&=\int_0^1\frac{a\ln\left(\frac{1-x}{x}\right)}{1+a^2x^2}\ dx\tag{**}
\end{align}
From (*) and (**) we have 

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


Note:
$-\frac{H_{2n+1}}{2n+1}=\int_0^1x^{2n}\ln(1-x)\ dx$
$\frac1{(2n+1)^2}=-\int_0^1 x^{2n}\ln x\ dx$
</s>