The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?

Another proof i have (re?)discovered.
I want to prove that,
$\displaystyle J:=\int_0^1 \frac{\ln(1+x)}{x}dx=\frac{\pi^2}{12}$
Let $f$, be a function, such that, for $s\in[0;1]$,
$\displaystyle f(s)=\int_0^{\frac{\pi}{2}} \arctan\left(\frac{\sin t}{\cos t+s}\right)\,dt$
Observe that,
$\begin{align} f(0)&=\int_0^{\frac{\pi}{2}}\arctan\left(\frac{\sin t}{\cos t}\right)\,dt\\
&=\int_0^{\frac{\pi}{2}} t\,dt\\
&=\left[\frac{t^2}{2}\right]_0^{\frac{\pi}{2}}\\
&=\frac{\pi^2}{8}
\end{align}$
For $t$ in $\left[0,\frac{\pi}{2}\right]$,
$\begin{align} \frac{\sin t}{\cos t+1}&=\frac{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)}{\cos^2\left(\frac{t}{2}\right)-\sin^2\left(\frac{t}{2}\right)+1}\\
&=\frac{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)}{2\cos^2\left(\frac{t}{2}\right)}\\
&=\tan\left(\frac{t}{2}\right)
\end{align}$
Therefore,
$\begin{align} f(1)&=\int_0^{\frac{\pi}{2}}\arctan\left(\frac{\sin t}{\cos t+1}\right)\,dt\\
&=\int_0^{\frac{\pi}{2}}\arctan\left(\tan\left(\frac{t}{2}\right)\right)\,dt\\
&=\int_0^{\frac{\pi}{2}} \frac{t}{2}\,dt\\
&=\left[\frac{t^2}{4}\right]_0^{\frac{\pi}{2}}\\
&=\frac{\pi^2}{16}
\end{align}$
For $s$ in $[0,1]$,
$\begin{align}
f^\prime(s)&=-\int_0^{\frac{\pi}{2}}\frac{\sin t}{1+2s\cos t+s^2}\,dt\\
&=\left[\frac{\ln(1+2s\cos t+s^2)}{2s}\right]_0^{\frac{\pi}{2}}\\
&=\frac{1}{2}\frac{\ln\left(1+s^2\right)}{s}-\frac{\ln\left(1+s\right)}{s}
\end{align}$
Therefore,
$\begin{align}
f(1)-f(0)&=\int_0^1 f^\prime(s)ds\\
&=\frac{1}{2}\int_0^1\frac{\ln\left(1+s^2\right)}{s}\,ds-\int_0^1 \frac{\ln\left(1+s\right)}{s}\,ds\\
\end{align}$
In the first integral perform the change of variable $y=s^2$, therefore,
$\displaystyle f(1)-f(0)=-\frac{3}{4}J$
But,
$\begin{align} f(1)-f(0)&=\frac{\pi^2}{16}-\frac{\pi^2}{8}\\
&=-\frac{\pi^2}{16}
\end{align}$
Therefore,
$\boxed{\displaystyle J=\frac{\pi^2}{12}}$
PS:
To obtain the value of $J$ knowing that $\displaystyle \zeta(2)=-\int_0^1 \frac{\ln(1-x)}{x}dx$
$\begin{align}
\int_0^1 \frac{\ln(1+t)}{t}\,dt+\int_0^1 \frac{\ln(1-t)}{t}\,dt=\int_0^1 \frac{\ln(1-t^2)}{t}\,dt
\end{align}$
Perform the change of variable $y=t^2$ in RHS integral,
$\begin{align}
\int_0^1 \frac{\ln(1+t)}{t}\,dt+\int_0^1 \frac{\ln(1-t)}{t}\,dt=\frac{1}{2}\int_0^1 \frac{\ln(1-t)}{t}\,dt
\end{align}$
Therefore,
$\begin{align}
\int_0^1 \frac{\ln(1+t)}{t}\,dt=-\frac{1}{2}\int_0^1 \frac{\ln(1-t)}{t}\,dt
\end{align}$
$\boxed{\displaystyle \int_0^1 \frac{\ln(1+t)}{t}\,dt=\frac{1}{2}\zeta(2)}$
