So, integral question time.
$$\int_0^1 \frac{1}{x} \ln\left(\frac{1+x}{1-x}\right)e^{-x}\ \text{d}x$$
I was trying to compute this integral, and at a certain point I checked with Mathematica, and it solved it in a numerical way, returning the following result:
$$1.4721301560626(...)$$
I am now asking if there is a way to obtain this result in a close form. I found out that that number can be expressed as
$$\frac{1}{8}\left(60 C - 4 - 21\pi +\pi^2 + 68\pi \ln(2) - 38\pi \ln(3)\right)$$
Where $C$ is the Catalan constant.
But even better as
$$\frac{1097113}{2341289}\pi$$
Despite those forms are not exact.
Is there a way to solve that integral in order to get the result in a close form?
I am not saying I strongly think there must be a close form, but it would be nice!
For any $k\geq 1$ we have $$ \int_{0}^{1}\log(1-x)\, x^{k-1} \,dx = -\frac{H_k}{k},\qquad \int_{0}^{1}\log(1-x^2)\, x^{k-1} \,dx = -\frac{H_{k/2}}{k} \tag{A}$$ hence it follows that $$ \int_{0}^{1}\log\left(\frac{1+x}{1-x}\right)e^{-x}\,\frac{dx}{x} = \frac{\pi^2}{4}+\!\!\!\!\!\!\underbrace{\sum_{k\geq 1}\frac{(-1)^k}{k!}\cdot \frac{2H_k-H_{k/2}}{k}}_{\text{fast-convergent and with alternating signs}}\tag{B}$$
and plenty of accurate approximations are simple to derive. On the other hand, I would not bet on the existence of a simple closed form for the highlighted series, if not in terms of exponential integrals.