How do we show $(1)$ is correct? $$\int_{0}^{1}\ln\left({1+x\over 1-x}\right){\mathrm dx\over 1+x^2}=C\tag1$$
Where C is the Catalan constant
$u={1+x\over 1-x}$ then $du={2dx\over (1-x)^2}$
$${1\over 2}\int_{1}^{\infty}(1-x)^2\ln(u){\mathrm du\over 1+x^2}=C\tag2$$
$x={u-1\over u+1}$
$${1\over 2}\int_{1}^{\infty}{4\over (u+1)^2}\ln(u){\mathrm (u+1)^2du\over 2(u^2+1)}=C\tag3$$
$$\int_{1}^{\infty}\ln(u){\mathrm du\over 1+u^2}=C\tag4$$
On
From what you did if you set $u = \frac{1}{t}$ that is $du = -\frac{1}{t^2} $ then We get
$$ C= \int_{1}^{\infty}\ln(u){\mathrm du\over 1+u^2}=\int_{0}^{1}\frac{\ln \frac{1}{t}}{1+\frac{1}{t^2}}\,\frac{1}{t^2}dt=-\int_{0}^{1}\frac{\ln t}{1+t^2}\,dt $$ That is $$ C= -\int_{0}^{1}\frac{\ln t}{1+t^2}\,dt=\int_{0}^{1}\sum_{n= 0}^{\infty}-(-1)^n t^{2n}\ln t dt \\=\int_{0}^{1}\sum_{n= 0}^{\infty}-(-1)^n t^{2n}\ln t dt \\ =\sum_{n= 0}^{\infty}\frac{(-1)^n}{(2n+1)^2}$$
Where we used that,
$$\frac{1}{1+x}= \sum_{n= 0}^{\infty} (-1)^nx^n ~~~|x|<1$$
and simple integration by part gives: $$\int_{0}^{1} t^{2n}\ln t dt= \frac{1}{2n+1} \left[t^{2n+1}\ln t\right]^1_0 -\frac{1}{2n+1} \int_{0}^{1} t^{2n} dt = -\frac{1}{(2n+1)^2} $$
Since $$\lim_{t\to 0 } t^{2n+1}\ln t = 0$$
As you noticed, it is enough to prove $(4)$. By enforcing the substitution $u\mapsto \frac{1}{v}$ we are left with
$$ \int_{0}^{1}\frac{-\log v}{1+v^2}\,dv \stackrel{?}{=}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2} $$ which is almost trivial since $\frac{1}{1+v^2}=1-v^2+v^4-v^6+\ldots$ for $v\in(0,1)$ and $$ \int_{0}^{1}v^{2k}\left(-\log v\right)\,dv = \frac{1}{(2k+1)^2} $$ is straightforward to prove through the substitution $v\mapsto e^{-z}$ and the $\Gamma$ function, or just by differentiation under the integral sign (Feynman's trick, see Chapter 4 of these notes).