Are we aware of an elementary way of proving that?

$$\int\_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=\frac{13}{24} \zeta (3)$$

Of course, with the help of Mathematica it can be done, but I wonder if there exists an elementary, simple, easy way of finishing it.

$$\int \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=$$
$$=\text{Li}\_3\left(\frac{t+2}{t}\right)-\text{Li}\_3\left(-\frac{t+2}{t}\right)+\text{Li}\_3(-t)+\text{Li}\_3(t+2)-\left(\text{Li}\_2\left(\frac{t+2}{t}\right)-\text{Li}\_2\left(-\frac{t+2}{t}\right)\right) \log \left(-\frac{t+2}{t}\right)-\text{Li}\_2(-t-1) \left(\log \left(\frac{1}{t}\right)+\log (t)\right)-\text{Li}\_2(-t) \left(\log (t+2)-\log \left(-\frac{t+2}{t}\right)\right)-\text{Li}\_2(t+2) \left(\log \left(-\frac{t+2}{t}\right)+\log (t)\right)-\frac{1}{2} \left(-\log \left(\frac{2 (t+1)}{t}\right)+\log \left(-\frac{2}{t}\right)+\log (t+1)\right) \log ^2\left(-\frac{t+2}{t}\right)-(\log (-t-1)-\log (t+1)) \log (t+2) \log \left(-\frac{t+2}{t}\right)-\log (t) \log (t+1) \log (t+2)-\frac{1}{2} (\log (t+1)-\log (-t-1)) \log (t+2) (\log (t+2)-2 \log (t))$$

that is obtain by Mathematica.

**EDIT:** I think once we rewrite all as

$$ \frac{1}{2}\int\_0^1 \left(\frac{\log ^2\left(\frac{t+2}{t}\right)}{t+1}-\frac{\log ^2(t+2)}{t+1}-\frac{\log ^2\left(\frac{1}{t}\right)}{t+1}\right) \, dt$$
and employ the proper variable change, we're almost done, the idea that just crossed my mind.

The hardest part is probably the integral 
$$\int\_0^1 \frac{\log ^2\left(\frac{t+2}{t}\right)}{t+1} \ dt$$ where 
by letting $t/(t+2)\mapsto t$, we obtain the far nicer form

$$2\int\_0^{1/3} \frac{\log ^2(t)}{1-t^2} \, dt$$
$$=\int\_0^{1/3} \frac{\log ^2(t)}{ 1-t} \ dt+\int\_0^{1/3}\frac{\log ^2(t)}{ 1+t} \, dt$$
where then using geometric series and swapping the integration and summation, we obtain that$$\int\_0^1 \frac{\log ^2\left(\frac{t+2}{t}\right)}{t+1} \ dt=$$
$$2\text{Li}\_3\left(\frac{1}{3}\right)-2\text{Li}\_3\left(-\frac{1}{3}\right)+ \log (3) \left(4 \text{Li}\_2\left(\frac{1}{3}\right)-\text{Li}\_2\left(\frac{1}{9}\right)+\log (2) \log (3)\right) \tag1$$
Using geometric series again, we easily get that 

$$\int\_0^1 \frac{\log ^2\left(\frac{1}{t}\right)}{t+1} \, dt=\frac{3}{2} \zeta (3) \tag2$$
The last integral can be reduced by variable change to a well-known integral treated by Leonard Lewin in his book about polylogarithms, that is $$\int\_0^1 \frac{\log ^2(t+2)}{t+1} \, dt=\int\_1^2 \frac{\log ^2(t+1)}{t} \, dt$$
So, we can use the fact that 
$$\int\_0^t \frac{\log ^2(1+t)}{t} \, dt$$
$$=\log(t)\log^2(1+t)-2/3\log^3(1+t)-2\log(1+t)\operatorname{Li}\_2\left(\frac{1}{1+t}\right)-2\operatorname{Li}\_3\left(\frac{1}{1+t}\right)+2\zeta(3)$$

So, using the foregoing result, we immediately get that 
$$\int\_1^2 \frac{\log ^2(t+1)}{t} \, dt$$
$$=-2 \text{Li}\_3\left(\frac{1}{3}\right)-2 \text{Li}\_2\left(\frac{1}{3}\right) \log (3)+\frac{7 \zeta (3)}{4}-\frac{1}{3} 2 \log ^3(3)+\log (2) \log ^2(3) \tag3$$

Combining $(1) (2) (3)$, we obtain that
$$\int\_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt$$
$$=2 \text{Li}\_3\left(\frac{1}{3}\right)-\text{Li}\_3\left(-\frac{1}{3}\right)-\frac{1}{2} \left(\text{Li}\_2\left(\frac{1}{9}\right)-6 \text{Li}\_2\left(\frac{1}{3}\right)\right) \log (3)-\frac{13 \zeta (3)}{8}+\frac{\log ^3(3)}{3}$$

Using $(20)$ from here <http://mathworld.wolfram.com/Dilogarithm.html> and then the identity and $(1)$ from this question/answer [Proving $\text{Li}\_3\left(-\frac{1}{3}\right)-2 \text{Li}\_3\left(\frac{1}{3}\right)= -\frac{\log^33}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}$?](https://math.stackexchange.com/questions/676124/proving-textli-3-left-frac13-right-2-textli-3-left-frac13-rig), we get that 

$$\int\_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=\frac{13}{24} \zeta (3).$$

**A SPECIAL ADDENDUM:**
The present integral can be viewed as the key core of calculating a pretty tough integral that was posted on MSE some time go, 
$$\int\_0^1 \frac{\text{Li}\_2 \left(-\frac{1}{1-z}\right)-\text{Li}\_2 \left(-\frac{1}{1+z}\right)}{z}dz$$ at this link [Evaluating $\int\_0^1 \frac{\text{Li}\_2 \left(-\frac{1}{1-z}\right)-\text{Li}\_2 \left(-\frac{1}{1+z}\right)}{z}dz$](https://math.stackexchange.com/questions/645974/evaluating-int-01-frac-textli-2-left-frac11-z-right-textli-2).

Simple integration by parts combined with the use of the geometric series shows that 

$$\int\_0^1 \frac{\text{Li}\_2 \left(-\frac{1}{1-z}\right)-\text{Li}\_2 \left(-\frac{1}{1+z}\right)}{z}dz$$ 
$$=\int\_0^1 \frac{\log (z+2) \log (z)}{z+1}dz+\underbrace{\int\_0^1\frac{\log (2-z) \log (z)}{1-z} \ dz}\_{\large \sum \_{n=1}^{\infty } \frac{(-1)^n H\_n}{n^2}=-5/8 \zeta (3)}-\underbrace{\int\_0^1\frac{\log (1-z) \log (z)}{1-z}dz}\_{\large\sum \_{n=1}^{\infty } \frac{H\_n}{(n+1)^2}=\zeta (3)}-\underbrace{\int\_0^1\frac{\log (z+1) \log (z)}{z+1} \, dz}\_{\large \sum \_{n=1}^{\infty } \frac{(-1)^n H\_n}{(n+1)^2}=-1/8\zeta (3)}$$
where since the integral unknown is calculated in this post, we conclude that 

$$\int\_0^1 \frac{\text{Li}\_2 \left(-\frac{1}{1-z}\right)-\text{Li}\_2 \left(-\frac{1}{1+z}\right)}{z}dz$$ 
$$=-\frac{49}{24}\zeta(3).$$
**A SPECIAL ADDENDUM $2$:** By a careful approach of the previous results, we also obtain the value of the very beautiful series 

$$\sum \_{k=1}^{\infty } \sum \_{n=1}^{\infty } \frac{(-1)^k }{k2^k n 2^n (k+n)}=\frac{1}{12} \left(3 \text{Li}\_3\left(\frac{1}{4}\right)+6\log (2) \text{Li}\_2\left(\frac{1}{4}\right) +4 \log ^3(2)-4 \zeta (3)\right)$$

**Q.E.D.**