A lucky proof for the Basel problem. 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.

Can someone prove rigorously that we are allowed to plug in $b=-1$ in order to get the correct result? 

 A: *

*In fact, the equation $$\int_0^{\infty}\frac{\ln x\mathrm{d}x}{(x+a)(x+b)}=\frac{(\ln a)^2-(\ln b)^2}{2(a-b)}$$ holds for all distinct complex $a$, $b$ not lying on the nonpositive real axis, as follows immediately from the contour integral representation $$\int_0^{\infty}\frac{\ln x\mathrm{d}x}{(x+a)(x+b)}=-\int_H\frac{\mathrm{d}x}{2\pi\mathrm{i}}\frac{(\ln x)^2}{2(x-a)(x-b)}\text{.}$$ Here $H$ is a Hankel contour about the negative real axis—where the branch cut of $\ln$ is chosen to be.

*Even if $b<0$, we may consider the Cauchy principal value $$\mathrm{P}\int_0^{\infty}\frac{\ln x\mathrm{d}x}{(x+a)(x-\lvert b\rvert)}\text{.}$$ In this case, we may use the Sokhotski–Plemelj theorem to arrive at $$\mathrm{P}\int_0^{\infty}\frac{\ln x\mathrm{d}x}{(x+a)(x-\lvert b\rvert)}=\frac{(\ln a)^2-(\ln \lvert b\rvert )^2+\pi^2}{2(a+\lvert b\rvert)}\text{.}$$

*If we let $b=-1$, then the integral converges and is equal to its Cauchy principal value. Therefore $$\int_0^{\infty}\frac{\ln x\mathrm{d}x}{(x+a)(x-1)}=\frac{(\ln a)^2+\pi^2}{2(a+1)}\text{.}$$
A: Here is an 'elementary' proof of its validity. Note that
$$ \frac{c+1}{(x+c)(x-1)}= \frac{c-1}{(x+c)(x+1)} +\frac{2}{x^2-1} $$
Then
$$ \int_0^\infty \frac{\ln x}{(x+c)(x-1)}dx = \frac{c-1}{c+1}I_1 + \frac{2}{c+1}I_2 \tag{1}$$
where
$$ I_1 =\int_0^\infty \frac{\ln x }{(x+c)(x+1)}
\overset{t=c/x}{dx}=\int_0^\infty \frac{\ln c }{(t+c)(t+1)}dt- I_1 = \frac{\ln^2 c}{2(c-1)}\tag{2}$$
and
$I_2 =\int_0^\infty \frac{\ln x}{x^2-1}dx$. Define
$J(\alpha)=\int_0^\infty \frac{\ln (1-\alpha^2 + \alpha^2 x^2)}{2(x^2-1)}dx $
\begin{align}
 J'(\alpha)=\int_0^\infty \frac{\alpha}{1-\alpha^2 + \alpha^2 x^2} dx = \frac{\pi/2}{\sqrt{1-\alpha^2}}
\end{align}
Then
$$ I_2 = J(1) = \int_0^1 J'(\alpha) d\alpha = \frac{\pi}{2}\int_0^1 \frac{1}{\sqrt{1-\alpha^2}} d\alpha = \frac{\pi^2}{4} \tag{3}$$
Substitute (2) and (3) into (1), we obtain the sought-after result
$$ \int_0^\infty \frac{\ln x}{(x+c)(x-1)}dx = \frac{\pi^2 + \ln^2 c}{2(c+1)}$$
A: Fixing $a >0$, we see that the integral
\begin{align*}
f(z) := \int_0^\infty \frac{\log x}{(x+a)(x+z)} dx.
\end{align*} converges absolutely for $z$ away from $(-\infty,0]$. So $f$ defines an analytic function on $\mathbb C \setminus (-\infty,0]$ (which can be checked by Morera's theorem and Fubini's theorem, for example.) This implies
$$
f(z) = \frac{\log^2 a - \log^2 z}{2(a-z)}
$$ holds not only on $(0,\infty)$ but also $\mathbb C \setminus (-\infty,0]$ by analytic continuation.
Now we need to check the continuity of $f$ at $z=-1$. Let 
$$
F(x) = \int_1^x \frac{ \log t}{(t+a)(t-1)} dt,\quad x\ge 0
$$ so that $F(\infty) - F(0) $ is the desired integral $\displaystyle I(a) = \int_0^\infty \frac{\log x}{(x+a)(x-1)} dx$. By integration by parts,
\begin{align*}
f(z) =& \left[F(x)\frac{ x-1}{x+z}\right]^\infty _0 - (1+z)\int_0^\infty \frac{F(x)}{(x+z)^2} dx\\
=& \left(F(\infty) +\frac{F(0)}{z} \right)- (1+z)\int_0^\infty \frac{F(x)}{(x+z)^2} dx,\quad z\neq 0
\end{align*} The first term tends to $I(a)$ as $z\to -1$. For $z=-1+it, t>0$, the second term can be estimated as
\begin{align*}
|1+z|\left|\int_0^\infty \frac{F(x)}{(x+z)^2} dx\right| \le& t \int_0^\infty \frac{|F(x)|}{|x-1+it|^2} dx \\
\le& \int_0^\infty \frac{t}{(x-1)^2 + t^2} |F(x)| dx\\
&\xrightarrow{t\to 0} \pi |F(1)| = 0.
\end{align*} Thus
$$
 I(a)= \lim_{t\to 0^+} f(-1+it) =\frac{\log^2 a +\pi^2}{2(1+a)} .
$$
Note that this argument only works for $z=-1$ since $\displaystyle F(x) =\int_c^x \frac{\log t}{(t+a)(t-c)} dt$ is not well-defined (due to the singularity at $t=c$) for $c>0, c\ne 1$.
