<s>[INST] Computing $\int_{-1}^{1} \frac{\ln (1+y)}{y} d y$ with simple method. Using the series
$$
\ln (1+y)=\sum_{n=0}^{\infty} \frac{(-1)^{n} y^{n+1}}{n+1} \text { for }|y|<1
$$
to convert the integral into $$
\begin{aligned}
\int_{-1}^{1} \frac{\ln (1+y)}{y} d y &=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} \int_{-1}^{1} y^{n} d y \\
&=\sum_{n=0}^{\infty} \frac{1-(-1)^{n+1}}{(n+1)^{2}} \\
&=\zeta(2)+\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n+1)^{2}} \\
&=\zeta(2)+\frac{1}{2}\zeta(2)\\&=\frac{\pi^{2}}{4}
\end{aligned}
$$
Question: Is there any other simple solution? Your suggestion and solution is warmly welcome.
 [/INST] An elementary solution
\begin{align}
\int_{-1}^{1} \frac{\ln (1+y)}{y} &{d y}
\overset{y\to -y} = \frac12 \int_{-1}^{1} \frac{\ln \frac{1+y}{1-y}}{y} d y
\overset{y\to \frac{1+y}{1-y}}=\int_0^\infty 
\frac{\ln y}{y^2-1}dy\\
=& \int_0^\infty \int_0^1 \frac{x}{1+(y^2-1)x^2}dx\>dy
=\frac\pi2 \int_0^1 \frac1{\sqrt{1-x^2}}dx=\frac{\pi^2}4
\end{align}
</s>