Integral from Mathematica's documentation: $\int_0^1 \frac{\log (\frac{1}{2}(1+\sqrt{4 x+1}))}{x} \, dx = \frac{\pi^2}{15} $ I like to peruse Mathematica's documentation and look at the 'Neat Examples': this is one I managed to figure out. Apparently it's due to Ramanujan:
$$
I=\int_0^1 \frac{\log \left(\frac{1}{2} \left(1+\sqrt{4 x+1}\right)\right)}{x}
   \, dx = \frac{\pi^2}{15}. 
$$

Here are the steps for my solution:

*

*Make the substitution $x=y^2-y$, yielding
$$
    I= \int _{1}^{\phi}\frac{\log(y)(2y-1)}{y(y-1)}\,dy,  
$$where $\displaystyle{\phi = \frac{1+\sqrt{5}}{2}}$ is the golden ratio.

*Factor out the $\log(y)$ term and use partial fractions to write
$$I = \underbrace{\int _{1}^{\phi}\frac{\log(y)}{y}\,dy}_{I_1}  + \underbrace{\int _{1}^{\phi}\frac{\log(y)}{y-1}\,dy}_{I_2}
$$$I_1$ can be evaluated using a simple substitution, yielding $\displaystyle{I_1 = \frac{\log ^2(\phi )}{2}}$.

*Use the Taylor series for $\log(y)$ centered at $y=1$ and interchange the sum and integral to show
$$
    I_2 = -\sum_{k=1}^{\infty} \frac{(1-\phi)^{k}}{k^2}= -\sum_{k=1}^{\infty} \frac{(-\phi^{-1})^{k}}{k^2}= - \text{Li}_2(-\phi^{-1})
$$

*$\text{Li}_2$ has the following properties:


*

*$\text{Li}_2(x) + \text{Li}_2(-x) = \frac{1}{2}\text{Li}_2(x^2)$

*$\text{Li}_2(x) + \text{Li}_2(1-x) = \zeta(2) - \log(x)\log(1-x)$

*$\text{Li}_2(1-x) + \text{Li}_2(1-x^{-1}) = -\frac{1}{2}\log^2(x)$
Put $x=\phi^{-1}$ and use $\phi^2=\phi+1$; this gives:
$$
    \text{Li}_2(\phi^{-1}) + \text{Li}_2(-\phi^{-1}) = \frac{1}{2}\text{Li}_2(1-\phi^{-1})
$$
$$
    \text{Li}_2(\phi^{-1}) + \text{Li}_2(1-\phi^{-1}) = \zeta(2) -2 \log^2(\phi)
$$
$$
    \text{Li}_2(1-\phi^{-1}) + \text{Li}_2(-\phi^{-1}) =-\frac{1}{2}\log^2(\phi)
$$
5. Relabel for clarity. Let $A=\text{Li}_2(\phi^{-1})$, $B=\text{Li}_2(-\phi^{-1})$, $C=\text{Li}_2(1-\phi^{-1})$, and $L= \log^2(\phi)$. This gives the system
$$
    \begin{cases}
    A+ B & = \frac{1}{2}C\\
    A+ C&= \zeta(2)- 2L\\
    C+B &= -\frac{1}{2}L
    \end{cases}
$$Solving gives $B=-I_2=\displaystyle{\frac{1}{2}L-\frac{2}{5}\zeta(2)}$, whence $\displaystyle{I = \frac{\pi^2}{15}}.$

I'd be curious to see if there are any other methods of proof, perhaps involving simpler substitutions than the ones I used.
 A: The integral can be generalized and solved in terms of dilogarithms:
$$ (1) \quad \int_0^1  \log{\big( (1+\sqrt{1+y\ x})/2\big)} \frac{dx}{x} = 
-\frac{1}{2}\Big\{ \zeta(2) + \text{Li}_2\big( (1-\sqrt{1+y})/2\big) + $$
$$ + \text{Li}_2\big( (1+\sqrt{1+y})/2\big)  - \log{(-y/4)} 
\log{\big( (1+\sqrt{1+y})/2\big)}
\big) \Big\}$$
Note that for $y=4,$ the analytic continuation of the dilogarithm must be used because the last two terms will have an imaginary part.  In the dilog term on the top line of (1), the argument is $\rho=1/2(\sqrt{5} - 1).$  The dilog at this argument and its square has been known since 1780 (long before Ramanujan) and is due to Landen:
$$ \text{Li}_2(\rho) = \pi^2/10 - \log^2\rho \ , \ \text{Li}_2(\rho^2) = \pi^2/15 - \log^2\rho $$
My proof of (1) ultimately depends on finding a close-form expression for
$$ (2) \quad \sum_{k=1}^\infty \frac{(-y/4)^k}{k^2}\binom{2k}{2} = \frac{1}{\pi}\int_0^1
\text{Li}_2(-x\,y)\frac{dx}{\sqrt{x(1-x)}}.$$
(The answer to both is within the curly braces of (1).)  I doubt my contribution is original.  Mathematica knows about Landen's formulas and plugging $y=4$ and letting it do the simplification gives $I=\pi^2/15.$
