A closed form indeed exists for this integral:
$$\int_0^\infty\frac{\log\left(1+\frac{\pi^2}{4\,x}\right)}{e^{\sqrt{x}}-1}\mathrm dx=\pi^2\left(\log\frac{\pi\,A^9\sqrt{2}}{\Gamma(\frac{1}{4})^2}-\frac{9}{8}\right)+2\,\pi\,C,$$
where $A$ is the Glaisher-Kinkelin constant and $C$ is the Catalan constant.
A more general result:
$$\int_0^\infty\frac{\log\left(1+\frac{a}{x}\right)}{e^{\sqrt{x}}-1}\mathrm dx=8\,\pi^2\psi^{(-2)}\left(\frac{\sqrt{a}}{2\pi}\right)-\frac{a}2\left(1+\log\frac{4\pi^2}{a}\right)-2\pi\sqrt{a}\left(1+2\log\Gamma\left(\frac{\sqrt{a}}{2\pi}\right)\right),$$
where $\psi^{(-2)}(z)$ is the generalized polygamma function.
The proof is based on Binet's second formula, but I still need to sort out some details.
With some help from Mathematica I got this result:
$$\int_0^\infty\left(5\,x^5+x\right)\operatorname{erfc}\left(x^5+x\right)dx=J_{\frac25}\left(\frac8{25\sqrt5}\right)\left(\frac8{375}\sqrt{\frac25\left(5-\sqrt5\right)}\,\pi\,J_{\frac45}\left(\frac8{25\sqrt5}\right)-\frac8{375}\sqrt{\frac25\left(5+\sqrt5\right)}\,\pi\,J_{\frac65}\left(\frac8{25\sqrt5}\right)\right)+J_{-\frac25}\left(\frac8{25\sqrt5}\right)\left(\frac1{75}\sqrt{2\left(5+\sqrt5\right)}\,\pi\,J_{-\frac15}\left(\frac8{25\sqrt5}\right)+\frac1{25}\left(\sqrt5-5\right)\sqrt{\frac1{10}\left(5+\sqrt5\right)}\,\pi\,J_{\frac15}\left(\frac8{25\sqrt5}\right)-\frac8{375}\sqrt{\frac25\left(5+\sqrt5\right)}\,\pi\,J_{\frac45}\left(\frac8{25\sqrt5}\right)-\frac{4\left(\sqrt5-5\right)\sqrt{2\left(5+\sqrt5\right)}\,\pi\,J_{\frac65}\left(\frac8{25\sqrt5}\right)}{1875}\right)+J_{-\frac15}\left(\frac8{25\sqrt5}\right)\left(\frac1{25}\sqrt{2\left(5-\sqrt5\right)}\,\pi\,J_{\frac25}\left(\frac8{25\sqrt5}\right)+\frac8{375}\sqrt{\frac25\left(5+\sqrt5\right)}\,\pi\,J_{\frac35}\left(\frac8{25\sqrt5}\right)-\frac8{375}\sqrt{\frac25\left(5-\sqrt5\right)}\,\pi\,J_{\frac75}\left(\frac8{25\sqrt5}\right)\right)+J_{\frac15}\left(\frac8{25\sqrt5}\right)\left(-\frac1{75}\sqrt{2\left(5+\sqrt5\right)}\,\pi\,J_{\frac25}\left(\frac8{25\sqrt5}\right)+\frac{4\left(\sqrt5-5\right)\sqrt{2\left(5+\sqrt5\right)}\,\pi\,J_{\frac35}\left(\frac8{25\sqrt5}\right)}{1875}+\frac8{375}\sqrt{\frac25\left(5+\sqrt5\right)}\,\pi\,J_{\frac75}\left(\frac8{25\sqrt5}\right)\right),$$
where $J_\nu(x)$ is the Bessel function of the first kind.
A solution outline:
Note that $(5\,x^5+x)=x\frac{d}{dx}(x^5+x)$. Change the integration variable $y=x^5+x$, then the integral takes the form
$$\int_0^\infty\mathcal{BR}(y)\cdot\operatorname{erfc}y\,dy,$$
where $y\mapsto\mathcal{BR}(y)$ is the inverse (properly selected to satisfy $\mathcal{BR}(y)>0$ for $y>0$) of the polynomial function $x \mapsto x^5+x$. This is a well-known non-elementary function called Bring radical, it can be used to express solutions of quintic equations in an explicit form.
An important fact, it has a representation via a generalized hypergeometric function (I used it to answer another question awhile ago). If we plug the hypergeometric representation into the integral and feed it to Mathematica, it produces the result in terms of Bessel functions shown above. I leave this step without a rigorous proof and rely on Mathematica here. The result agrees with a numerical integration to a very high precision. I would be very glad if anybody could write down an explicit derivation of the formula.
Best Answer
Here is a partial progress report. I am basically repeating Jim Belk's analysis from the previous answer.
Set $F(a) = \int_{x=0}^1 \frac{\log(1+x^a)}{1+x} dx$. Then $$F(a) = \int_{x=0}^1 \int_{y=0}^{x^a} \frac{dx dy}{(1+x)(1+y)} = \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)}$$ so $$F(a) + F(a^{-1}) = \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)} + \int_{0 \leq y \leq x^{1/a} \leq 1} \frac{dx dy}{(1+x)(1+y)}$$ $$= \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)} + \int_{0 \leq y^a \leq x \leq 1} \frac{dx dy}{(1+x)(1+y)} = \int_{0 \leq x,y \leq 1} \frac{dx dy}{(1+x)(1+y)} = (\log 2)^2.$$ (In order to combine the integrals, first switch the names of $x$ and $y$ in the second one.)
So $$F'(a) - a^{-2} F'(a^{-1})=0.$$ This gives a linear relation between $F'(2 + \sqrt{3})$ and $F'(2-\sqrt{3})$. If we find a second one, we can solve the linear equations and be done.
Notice that $$F'(a) = \int_{x=0}^1 \frac{x^a \log x dx}{(1+x)(1+x^a)} = \sum_{m,n=0}^{\infty} \int_{x=0}^1 (-1)^{m+n} x^{m+(n+1) a} \log x dx.$$ Integrating by parts, $\int_{x=0}^1 x^b \log x dx = \frac{-1}{(b+1)^2}$. So, ignoring issues of convergence, we should have $$F'(a) = \sum_{m,n=0}^{\infty} \frac{(-1)^{m+n+1}}{(m+(n+1) a + 1)^2} = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1}}{(m+n a)^2}$$ In the last step, we turned $m+1$ and $n+1$ into $m$ and $n$ to make things pretty. My guess is that the convergence issues can be dealt with for any $a>0$, but I haven't thought much about it.
So $$F'(a) + F'(a^{-1}) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left( \frac{(-1)^{m+n+1}}{(m+n a)^2} +\frac{(-1)^{m+n+1}}{(m+n a^{-1})^2} \right).$$ Putting $a=2 + \sqrt{3}$, this is $$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{m+n+1} \frac{2 (m^2+4mn+7n^2)}{(m^2+4mn+n^2)^2} $$ $$= 2 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1}}{m^2+4mn+n^2} +12 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1} n^2}{(m^2+4mn+n^2)^2}.$$
Here is where I run out of ideas. The first sum is basically the one at the end of Jim Belk's post, but I have no ideas for the second one.