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I'll modify this part since I want the proof to be here.

$$\sum_{n=1}^\infty \frac{1}{n^2}=\frac43\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=-\frac43\sum_{n=0}^\infty \int_0^1 x^{2n}\ln x dx=\frac43\int_0^1 \frac{\ln x}{x^2-1}dx$$ $$\int_0^1 \frac{\ln x}{x^2-1}dx\overset{x\rightarrow \frac{1}{x}}=\int_1^\infty \frac{\ln x}{x^2-1}dx\Rightarrow \sum_{n=1}^\infty \frac{1}{n^2}=\frac23 \int_0^\infty \frac{\ln x}{(x+1)(x-1)}dx$$

$$=\frac23 I(1,-1)=\frac23 \frac{\ln^2 (1)-\ln^2(-1)}{2(1-(-1))}=\frac23 \frac{\pi^2}{4}=\frac{\pi^2}{6}$$

Where we considered the following integral: $$I(a,b)=\int_0^\infty \frac{\ln x}{(x+a)(x+b)}dx\overset{x\rightarrow \frac{ab}{x}}=\int_0^\infty \frac{\ln\left(\frac{ab}{x}\right)}{(x+a)(x+b)}dx$$ Summing up the two integrals from above gives: $$2I(a,b)=\ln(ab)\int_0^\infty \frac{1}{(x+a)(x+b)}dx=\frac{\ln(ab)}{a-b}\ln\left(\frac{x+b}{x+a}\right)\bigg|_0^\infty $$ $$\Rightarrow I(a,b)=\frac{\ln(ab)}{2}\frac{\ln\left(\frac{a}{b}\right)}{a-b}=\frac{\ln^2 a-\ln^2 b}{2(a-b)}$$

From here we know that:

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

Also by plugging $b=-1$ in $I(a,b)$ we get: $$I(a,-1)=\int_0^\infty \frac{\ln x}{(x+a)(x-1)}dx=\frac{\ln^2a -\ln^2 (-1)}{2(a-(-1))}=\frac{\ln^2 a+\pi^2 }{2(a+1)}$$

We already know that this is true from the linked post, but let's ignore it, since the linked post uses the Basel problem to prove the result.