Evaluating $\int_0^1 \frac{\text{Li}_2 \left(-\frac{1}{1-z}\right)-\text{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz$ I was trying to find a closed form for
$$\int_0^1 \frac{\text{Li}_2 \left(-\frac{1}{1-z}\right)-\text{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz = -2.454199511\cdots$$
where $\text{Li}_2(z)$ is the dilogarithm function. Numerically, it seems very close to $-\frac{49}{24}\zeta(3)$.
How can we prove that $-\frac{49}{24}\zeta(3)$ is the exact value of the integral?
 A: Integrating by parts and using the known series results, we get that
$$\int_0^1 \frac{\text{Li}_2 \left(-\frac{1}{1-z}\right)-\text{Li}_2 \left(-\frac{1}{1+z}\right)}{z}dz$$ 
$$=\int_0^1 \frac{\log (z+2) \log (z)}{z+1}dz+\underbrace{\int_0^1\frac{\log (2-z) \log (z)}{1-z} dz}_{\large \sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^2}=-5/8  \zeta (3)}-\underbrace{\int_0^1\frac{\log (1-z) \log (z)}{1-z}dz}_{\large\sum _{n=1}^{\infty } \frac{H_n}{(n+1)^2}=\zeta (3)}-\underbrace{\int_0^1\frac{\log (z+1) \log (z)}{z+1} \, dz}_{\large \sum _{n=1}^{\infty } \frac{(-1)^n H_n}{(n+1)^2}=-1/8\zeta (3)}$$
For evaluation of the first integral, check my answer here Proving that $\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=\frac{13}{24} \zeta (3)$.
A: Wolframalpha can anti-differentiate your integrand. There are probably branch cuts involved which would make computing the integral more complicated than just plugging numbers into (or taking limits of) the anti-derivative.
A: I have been able to compute the integral analytically. The result is quite complex and I have not been able to simplify it. What I obtained is     
-Log[3]^3 / 3 + Log[3] PolyLog[2, 1/9] / 2 -  Log[27] PolyLog[2, 1/3] + PolyLog[3, -1/3] -
 2 PolyLog[3, 1/3] +  Zeta[3] / 8 
I hope and wish that there is no mistake.   
Generating the first 10,000 digits, the number is identical to - 49 Zeta[3] / 24 but I cannot prove it is true.  
