EDIT 12.02.2019
Inspired by the comment of omegadot "Based on the numerical evidence it seems correct with the final two expressions being far more complicated than I could have ever imagined. I am guessing it would also be an almighty herculean task to finally show $p_2-p_1 = \frac{\pi^2}{5\sqrt{5}}$"
I have solved both problems (without being Hercules).
First I showed, starting from my expressions $(2c)$ and $(3c)$, that indeed
$$p_2-p_1 = \frac{\pi^2}{5\sqrt{5}}\tag{e1}$$
Then simplifying $p_1$ using the identity mentioned in my comment
$$\text{Li}_2\left(\frac{1}{z}\right)+\text{Li}_2(z)+\frac{1}{2} \log ^2(-z)+\zeta (2)=0, z\lt 0\tag{e2}$$
and using $(e1)$ for $p_2$ I find with
$$q=\frac{1}{2} \left(\sqrt{5}-1\right)=\frac{1}{\phi}\tag{e3}$$
these simplified expressions (using $q$ or the golden ratio $\phi$, alternatively)
$$p_1 = \frac{4}{\sqrt{5}} \;\text{Li}_2\left(-q^2\right)+\frac{4 \log (q) \log (5 q)}{\sqrt{5}}+\frac{\pi ^2}{3 \sqrt{5}}\tag{2e}$$
$$p_1 = -\frac{4 \text{Li}_2\left(-\phi ^2\right)}{\sqrt{5}}-\frac{4 \log (\phi ) \log (5 \phi )}{\sqrt{5}}-\frac{\pi ^2}{3 \sqrt{5}}\tag{2f}$$
$$p_2 = \frac{4}{\sqrt{5}} \;\text{Li}_2\left(-q^2\right)+\frac{4 \log (q) \log (5 q)}{\sqrt{5}}+\frac{8 \pi ^2}{15 \sqrt{5}}\tag{3e}$$
$$p_2 = -\frac{4 \text{Li}_2\left(-\phi ^2\right)}{\sqrt{5}}-\frac{4 \log (\phi ) \log (5 \phi )}{\sqrt{5}}-\frac{2 \pi ^2}{15 \sqrt{5}}\tag{3f}$$
Remark
Observing that the parameter $q$ solves the equation $x^2+x-1=0$ led me to study the integral
$$i_2= \int_0^1 \frac{\log(x)}{x^2+x-1}\,dx\tag{e4}$$
which is similar to the integral
$$i_1= \int_0^1 \frac{\log(x)}{x^2-x-1}\,dx$$
which was the starting point in the post of omegadot.
I solved the (divergent) integral $i_2$ with a limiting procedure starting from a general polynomial of 2nd degree in the denominator which gives the result
$$i_2 = \frac{3 \pi ^2}{10 \sqrt{5}}+\frac{i \pi \log (q)}{\sqrt{5}}\tag{e5}$$
The real part is the principal value of $i_2$.
Note that this expression resembles the constant terms in $p_{1,2}$.
Original post, 11.02.2019
The question was "Is it possible to find closed-form expressions for the following two Euler sums containing the reciprocal of the central binomial coefficient?"
The answer is yes, and we provide both closed expressions.
These are given in both cases in terms of combinations of $\log(.)$ and the polylog function $\text{Li}_2(.)$. The arguments are expressions of the type $a + b \sqrt{5}$ with rational $a$ and $b$ where $\sqrt{5}$ can also be expressed in terms of the golden ratio.
Definitions
$$p_1= \sum_{n = 0}^\infty \frac{(-1)^n H_n}{(2n + 1) \binom{2n}{n}} \qquad \text{and} \qquad p_2= \sum_{n = 0}^\infty \frac{(-1)^n H_{2n + 1}}{(2n + 1) \binom{2n}{n}}$$
Here $H_n$ is the $n$th harmonic number.
Results
Using
$$H_{n}=\int_0^1 \frac{1-x^n}{1-x} \, dx\tag{1}$$
I find
For $p_1$
The sum under the integral is
$$s_1 = \frac{4 \left(\sqrt{5} \sqrt{x} \sqrt{x+4}\; \text{arcsinh}\left(\frac{1}{2}\right)-5\; \text{arcsinh}\left(\frac{\sqrt{x}}{2}\right)\right)}{5 (1-x) \sqrt{x} \sqrt{x+4}}\tag{2a}$$
We have to evaluate the integral
$$p_1 = \int_0^1 s_1 \,dx \tag{2b}$$
This can also be done in Mathematica to give
$$p_1 =-\frac{2}{\sqrt{5}} \left(\text{Li}_2\left(\frac{1}{2} \left(-3-\sqrt{5}\right)\right)-\text{Li}_2\left(\frac{1}{2} \left(-3+\sqrt{5}\right)\right)\tag{2c}\\+\log (25) \log \left(\frac{1}{2}+\frac{\sqrt{5}}{2}\right)\right)$$
Numerically,
$$N(p_1) \simeq -0.12711517793300195$$
Notice that $p_1$ can also be expressed by the "golden ratio
$$\phi=\frac{1}{2}(1+\sqrt{5}) \simeq 1.618033988749895$$
as
$$p_1=-\frac{2}{\sqrt{5}} (-\text{Li}_2(\phi -2)+\text{Li}_2(-\phi -1)+4 \log (2 \phi -1) \log (\phi ))\tag{2d}$$
For $p_2$
the sum can be done
$$s_2 = \frac{1}{1-x} \sum_{n = 0}^\infty (-1)^n (1-x^{2n + 1})\frac{1}{(2n + 1) \binom{2n}{n}}\\= \frac{4 \left(\sqrt{5} \sqrt{x^2+4} \;\text{arcsinh}\left(\frac{1}{2}\right)-5 \; \text{arcsinh}\left(\frac{x}{2}\right)\right)}{5 (1-x) \sqrt{x^2+4}}\tag{3a}$$
We have to evaluate the integral.
$$p_2 = \int_0^1 s_2 \,dx\tag{3b}$$
Mathematica 8.0 can do the integral (version 10.1 can't, so it is advisable to keep older versions).
The result (obtained after long series of surprising simplifications)
$$p_2 = \frac{4 \text{Li}_2\left(\frac{7}{2}-\frac{3 \sqrt{5}}{2}\right)}{\sqrt{5}}-\frac{4 \text{Li}_2\left(\frac{1}{2} \left(-3+\sqrt{5}\right)\right)}{\sqrt{5}}+\frac{2 \log \left(\frac{1}{2} \left(\sqrt{5}-1\right)\right) \log \left(25 \left(9-4 \sqrt{5}\right)\right)}{\sqrt{5}}\tag{3c}$$
Numerically,
$$N(p_2) \simeq 0.7556490761416742$$
In Terms of the golden ratio we have
$$p_2 = \frac{2}{\sqrt{5}}(\log (\phi -1) \log (25 (13-8 \phi ))-2 \text{Li}_2(\phi -2)+2 \text{Li}_2(5-3 \phi ))\tag{3d}$$
Technical remarks
Doing the integrals in Mathematica was by no means trivial. I proceeded in the common manner to calculate first the antiderivative and then - after having numerically confirmed continuity - take the limits at $x=0$ and $x=1$, resp.
While Version 10.1 refused to calculate the Limits at $x=1$, Version 8.0 provided results but they contained spurious imaginary parts and terms with products of concealed $0 * \infty$ so that the command "Simplify" returned "Indeterminate". A heuristic mixture of numerical checks and tetative cancelleations finally led me to the expressions given which were finally justified numerically.
We proved in this solution that
$$\mathcal{I}=\int_0^1\frac{\tan^{-1}(x)\ln(1+x^2)}{x(1+x)}dx=\frac{\pi^3}{96}-\frac{\pi}{8}\ln^2(2)\tag1$$
and we proved here that
$$\tan^{-1}x\ln(1+x^2)=-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}x^{2n+1}\tag2$$
By $(2)$ we get
$$\mathcal{I}=-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}\int_0^1\frac{ x^{2n}}{1+x}dx$$
Using the identity $$\int_0^1\frac{x^{2n}}{1+x}dx=\ln2+H_n-H_{2n}$$
it follows that
$$\mathcal{I}=-2\ln(2)\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}+2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{2}}{2n+1}$$
$$=-2\ln(2)\mathcal{S}_1-2\mathcal{S}_2+2\mathcal{S}_3\tag3$$
For $\mathcal{S}_1$ and $\mathcal{S}_3$, we use the classical identity:
$$\sum_{n=1}^\infty(-1)^n f(2n)=\Re\sum_{n=1}^\infty i^n f(n)$$
Therefore
$$\mathcal{S}_1=\Re\sum_{n=1}^\infty i^n\frac{H_n}{n+1}=\Re\left\{\frac{\ln^2(1-i)}{i}\right\}=-\frac{\pi}{8}\ln(2)\tag4$$
where we used $\sum_{n=1}^\infty x^n\frac{H_n}{n+1}=\frac{\ln^2(1-x)}{x}$ which follows from integrating $\sum_{n=1}^\infty H_nx^n=-\frac{\ln(1-x)}{1-x}$.
Similarly,
$$\mathcal{S}_3=\Re\sum_{n=1}^\infty i^n\frac{H_n^2}{n+1}$$
Using the generating function
$$\sum_{n=1}^\infty x^{n}\frac{ H_n^{2}}{n+1}=\frac{6\operatorname{Li}_3(1-x)-3\operatorname{Li}_2(1-x)\ln(1-x)-\ln^3(1-x)-3\zeta(2)\ln(1-x)-6\zeta(3)}{3x}$$
it follows that
$$\mathcal{S}_3=\Re\left\{\frac{6\operatorname{Li}_3(1-i)-3\operatorname{Li}_2(1-i)\ln(1-i)-\ln^3(1-i)-3\zeta(2)\ln(1-i)-6\zeta(3)}{3i}\right\}$$
$$=\Im\left\{\frac{6\operatorname{Li}_3(1-i)-3\operatorname{Li}_2(1-i)\ln(1-i)-\ln^3(1-i)-3\zeta(2)\ln(1-i)-6\zeta(3)}{3}\right\}$$
$$=2\Im\{\operatorname{Li}_3(1-i)\}+\frac12\ln(2)\ G+\frac{3\pi}{16}\ln^2(2)+\frac{5}{96}\pi^3\tag5$$
Plug the results of $(4)$ and $(5)$ in $(3)$ we get
$$\mathcal{I}=4\Im\{\operatorname{Li}_3(1-i)\}+\ln(2)\ G+\frac{5\pi}{8}\ln^2(2)+\frac{5}{48}\pi^3-2\mathcal{S}_2\tag6$$
By $(1)$ and $(6)$ we get
$$\mathcal{S}_2=\sum_{n=1}^\infty\frac{(-1)^nH_{2n}H_n}{2n+1}=2\Im\{\operatorname{Li}_3(1-i)\}+\frac12\ln(2)\ G+\frac{3\pi}{8}\ln^2(2)+\frac{3}{64}\pi^3$$
Best Answer
Similarly to this post. We have: $$4ab=(a+b)^2-(a-b)^2$$ Thus we can take $a=\ln(1-x)$ and $b=\ln(1+x)$ to get: $$I=\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x}dx=\frac14 \int_0^1 \frac{\ln^2(1-x^2)}{x}dx-\frac14 \int_0^1 \frac{\ln^2\left(\frac{1-x}{1+x}\right)}{x}dx$$ By setting $x^2 =t$ in the first integral and $\frac{1-x}{1+x}=t$ in the second one we get: $$I=\frac18 \int_0^1 \frac{\ln^2(1-t)}{t}dt -\frac12 \int_0^1 \frac{\ln^2 t}{1-t^2}dt=\frac18 \int_0^1 \frac{\ln^2t}{1-t}dt-\frac12 \int_0^1 \frac{\ln^2 t}{1-t^2}dt$$ $$=\frac18\sum_{n\ge 0}\int_0^1 t^n \ln^2 tdt-\frac12 \sum_{n\ge 0}\int_0^1 t^{2n}\ln^2 tdt$$ Using the following fact: $$\int_0^1 t^a dt=\frac{1}{a+1}\Rightarrow \int_0^1 t^a \ln^k tdt=\frac{d^k}{da^k} \left(\frac{1}{a+1}\right)=\frac{(-1)^k k!}{(a+1)^{k+1}}$$ $$\Rightarrow I=\frac18\sum_{n\ge 0}\frac{2}{(n+1)^3} -\frac12 \sum_{n\ge 0}\frac{2}{(2n+1)^3}=\frac28 \zeta(3) -\frac78 \zeta(3)=-\frac58\zeta(3)$$