Definite integral problem I was solving a definite integral problem which was reduced to :
$$\int^{1}_{0} \frac{\ln(1+t)}{t} dt$$
I couldn't solve it and when I saw the solution, the answer was simply given as $\frac{\pi^2}{12}$, and claimed that this is an identity. 
Can anybody give me a proof of this identity?

\begin{align*}
\int_0^1 \frac{\log(x+1)}{x} \, dx &= \int_0^1 \sum_{n=1}^\infty (-1)^{n+1} \frac{x^{n-1}}{n}dx\\ 
 &=  \sum_{n=1}^\infty (-1)^{n+1}\int_0^1 \frac{x^{n-1}}{n}dx\\ 
 &= \sum_{n=1}^\infty (-1)^{n+1} \frac{1}{n^2}\\ 
 &= \frac{\pi^2}{12}\\ 
\end{align*}
To calculate that sum, let us assume that the value of the following series
$$S_n = 1 + \frac 1 {2^2}  + \frac 1 {3^2} + \dots = \frac{\pi^2}{6}$$
Now if we consider only the even values, 
\begin{align*}
S_{2n} &= \frac 1 {2^2}  + \frac 1 {4^2} + \frac 1 {6^2} + \dots  \\ 
 &= \frac{1}{2^2 \cdot 1} + \frac 1 {2^2\cdot 2^2} + \frac{1}{2^2 \cdot 3^2} + \dots \\ 
 &= \frac{1}{4} S_n\\ 
 &= \frac{\pi^2}{24}\\ 
\end{align*}
To get the value of our series, we take $S_n - 2 S_{2n}$.
