Use $\sum_{n=1}^\infty \left( \frac{1}{n+a-1} - \frac{1}{n}\right) = -\gamma - \Psi(a)$, where $\Psi(a)$ is di-gamma function. Using this representation for the di-gamma function and integrating term-wise:
$ \begin{multline} \mathcal{I} = \int_0^\infty \frac{ \Psi(\frac{1}{4} + i \frac{x}{2}) + \Psi(\frac{1}{4} - i \frac{x}{2}) }{x^2 + 1/4} \, \mathrm{d} x = \\ \int_0^\infty \frac{ -2 \gamma }{x^2 + 1/4} \, \mathrm{d} x - \sum_{n=1}^\infty \int_0^\infty \frac{1}{x^2 + 1/4} \left( \frac{8 (4 n-3)}{(3-4 n)^2+4 x^2}-\frac{2}{n} \right) \, \mathrm{d} x \end{multline} $ Thus $ \mathcal{I} = -2 \pi \gamma - \sum_{n=1}^\infty \frac{2 \pi }{n (2 n-1)} = -2 \pi \left( \gamma + 2 \log 2 \right) = 2\pi \left( \Psi\left(\frac{3}{2}\right) - 2\right) $
Verify numerically:
In[55]:= NIntegrate[( PolyGamma[1/4 + I x/2] + PolyGamma[1/4 - I x/2])/( x^2 + 1/4), {x, 0, \[Infinity]}, WorkingPrecision -> 50] == 2 Pi (PolyGamma[3/2] - 2) Out[55]= True