Evaluate  $\int_0^1 {\ln(1+x)\over x}\,dx$. How would one evaluate   $\int_0^1 {\ln(1+x)\over x}\,dx$?
I'd like to do this without approximations. Not quite sure where to start. What really bothers me is that I came across this while reviewing my old intro to calculus book... but I'm fairly certain I've exhausted all the basic methods they teach in that text.
 A: $$ \int^1_0 \frac{ \log (1+x) }{x} dx = \int^1_0 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n-1}}{n} dx$$ 
$$ =\sum_{n=1}^{\infty} (-1)^{n-1} \int^1_0 \frac{x^{n-1} }{n} dx = \sum_{n=1}^{\infty} (-1)^{n-1}  \frac{1}{n^2}. $$
Denote $\displaystyle S = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.$ Then $$\sum_{n=1}^{\infty} (-1)^{n-1}  \frac{1}{n^2} = S - 2\left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \cdots \right) = S - \frac{S}{2} = \frac{\pi^2}{12}.$$
Thus $$\int^1_0 \frac{ \log (1+x) }{x} dx = \frac{\pi^2}{12}.$$
A: $$
I = \int_0^1 \frac{\ln(1+x)}{x}\,dx = \int_0^\infty \ln(1+e^{-t})\,dt\,,
$$
where $x = e^{-t}$.  Then expand
$$
\ln(1 + e^{-t}) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\,e^{-tn}\,,
$$
So we find
$$
I = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\,\int_0^\infty e^{-tn}\,dt = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} = \frac{\pi^2}{12}\,.
$$
Is the last sum familiar to you?  What about the closely related and easier sum
$$
\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}\,?
$$
A: The Taylor series for $\frac{\ln(1+x)}{x}$ is $\sum\limits_{n=1}^\infty (-1)^{n-1}\frac{x^{n-1}}{n}$, and this converges absolutely in $(0,1)$ thus we can use it for our integral. This means 
$$\int_0^1\frac{\ln(1+x)}{x}=\int_0^1\sum\limits_{n=1}^\infty (-1)^{n-1}\frac{x^{n-1}}{n}=\sum\limits_{n=1}^\infty (-1)^{n-1}\int_0^1\frac{x^{n-1}}{n}=\sum\limits_{n=1}^\infty (-1)^{n-1}\frac{1}{n^2}$$
and this series is equal to $\pi^2/12$ according to Wolfram|Alpha.
