Here there is an integral that I've found playing with Wolfram Alpha online calculator (thus to me is a curiosity that it has indefinite integral) $$\int\frac{\operatorname{Li}_2(x)}{1+\sqrt{x}}\,dx,\tag{1}$$ where the function in the numerator of the integrand is the polylogarithm
$$\operatorname{Li}_2(z)=\sum_{k=1}^\infty\frac{z^k}{k^2},\tag{2}$$ see the related MathWorld's article if you want to know more about this special function.
Question. To me seem difficult to get the indefinite integral that provide us Wolfram Alpha online calculator. But can you provide me the ideas or first calculations to calculate such indefinite integral? That is, imagine that I need to explain to a friend/colleague a draft about the strategy to get such indefinite integral. Then, what is the recipe that I need to explain him/her to justify from the top (without all tedious details) the indefinite integral? Many thanks.
When I was playing, before knowing such a closed form of the indefinite integral, my intention was to justify $$\int_0^1\frac{\operatorname{Li}_2(x)}{1+\sqrt{x}}\,dx.$$
I say these words to provide what are my intentions, I believe that this definite integral isn't special but I was interested in calculate it when I was asking to the mentioned CAS.
A natural temptation is to remove the square root from the denominator of the integrand function by enforcing the substitution $x=u^2$, then expanding $\text{Li}_2(u^2)$ as a Maclaurin series and convert the whole thing into a combination of Euler sums, hopefully with a low weight. Indeed
$$ \int_{0}^{1}\frac{\text{Li}_2(x)}{1+\sqrt{x}}=2\int_{0}^{1}\frac{u}{1+u}\text{Li}_2(u^2)\,du =4\int_{0}^{1}\left[1-\frac{1}{1+u}\right]\cdot\left[\text{Li}_2(u)+\text{Li}_2(-u)\right]\,du$$ where $$ \int_{0}^{1}\text{Li}_2(u^2)\,du = -4+\frac{\pi^2}{6}+4\log(2)$$ is straightforward and $$ \int_{0}^{1}\frac{\text{Li}_2(u)}{1+u}\,du =\frac{\pi^2}{6}\log(2)-\frac{5}{8}\zeta(3),\qquad \int_{0}^{1}\frac{\text{Li}_2(-u)}{1+u}\,du =-\frac{\pi^2}{12}\log(2)+\frac{1}{4}\zeta(3)$$ have already been proved on MSE. The involved techniques are just integration by parts and the functional relations for the dilogarithm function (a function with the sense of humour, according to D.Zagier).