Never seen all of these together

The function $(1+4x^2)^{-1} - \cos 2x$ vanishes at $x=0$ and has a well -defined Maclaurin series, so the whole integrand is continuous. There is therefore no problem integrating all the way down to $0$ .

The function $x^{-1}(1+4x^2)^{-1}$ can be integrated up to $\infty$ . The only problem is integrating $\frac{\cos 2x}{x}$ out to $\infty$ . But this is a standard improper Riemann integral.

On the other hand, it does not exist as a Lebesgue integral.

It all depends on how elementary you want to go! I will assume the following formula for the Euler-Mascheroni constant: $\gamma \; = \; -\int_0^\infty e^{-x}\, \ln x\,dx \;.$ Define functions $a,b,c$ on $(0,\infty)$ by $a(x) \; = \; \ln x \qquad b(x) \; = \; \int_0^x \frac{\cos t - 1}{t}\,dt \qquad c(x) \; = \; \int_x^\infty \frac{\cos t}{t}\,dt \;.$ Taking Laplace transforms of each of these functions ( $p > 0$ throughout): $\begin{array}{rcl} (La)(p) & = & \displaystyle \int_0^\infty \ln x e^{-px}\,dx \; = \; \int_0^\infty \ln\tfrac{x}{p} e^{-x} \,\tfrac{1}{p}dx \\ & = & \tfrac{1}{p} \displaystyle\int_0^\infty (\ln x - \ln p) e^{-x}\,dx \; = \; -\gamma p^{-1} - \tfrac{1}{p}\ln p \\ (Lb)(p) & = & \displaystyle \int_0^\infty\left(\int_0^x \frac{\cos t - 1}{t}\,dt\right) e^{-px}\,dx\; = \; \int_0^\infty dt \int_t^\infty \frac{\cos t - 1}{t} e^{-px}\,dx \\ & = & \displaystyle \tfrac{1}{p} \int_0^\infty \frac{\cos t - 1}{t}e^{-pt}\,dt \; = \; -\tfrac{1}{p}\int_0^\infty \left(\int_0^1 \sin st\,ds\right) e^{-pt}\,dt \\ & = & \displaystyle -\tfrac{1}{p} \int_0^1 ds \int_0^\infty e^{-pt}\sin st\,dt \; = \; -\tfrac{1}{p}\int_0^1 \frac{s}{p^2 + s^2}\,ds \\ & = & \displaystyle \Big[-\tfrac{1}{2p}\ln(p^2 + s^2)\Big]_0^1 \; = \; -\tfrac{1}{2p}\ln(p^2 + 1) + \tfrac{1}{p}\ln p \;. \end{array}$ Since $\tfrac{d}{dp}\int_0^\infty \frac{\cos t}{t} (1 - e^{-pt})\,dt \; = \; \int_0^\infty e^{-pt}\cos t\,dt \; = \; \tfrac{p}{p^2+1}$ we see that $\begin{array}{rcl} (Lc)(p) & = & \displaystyle\int_0^\infty \left( \int_x^\infty \frac{\cos t}{t}\,dt \right) e^{-px}\,dx \; = \; \int_0^\infty dt \int_0^t \frac{\cos t}{t} e^{-px}\,dx \\ & = & \displaystyle \tfrac{1}{p}\int_0^\infty \frac{\cos t}{t}(1 - e^{-pt})\,dt \; = \; \tfrac{1}{2p}\ln(p^2+1) \;. \end{array}$ Putting these results together yields $(La)(p) + (Lb)(p) + (Lc)(p) + \gamma p^{-1} \; \equiv \; 0$ and hence that $a(x) + b(x) + c(x) + \gamma \; \equiv \; 0$ This improves an important result about the Cosine Integral function $Ci(x) \; = \; -\int_x^\infty \frac{\cos t}{t}\,dt \; = \; -c(x) \; =\; a(x) + b(x) + \gamma \; = \; \gamma + \ln x + \int_0^x \frac{\cos t - 1}{t}\,dt$ This is the equation we need. This formula is frequently stated without proof; the above gives a sketch of why it is true (there is a lot of integration theory that is required to justify all the changes of integral order, and the differentiation within the integrals that I have done!).

We can now solve the problem! Since (by partial fractions) $\int_\epsilon^N \left(\frac{1}{1+4x^2} - \cos(2x)\right)\frac{dx}{x} \; = \; \int_\epsilon ^N \left(\frac{1 - \cos(2x)}{x} - \frac{4x}{1 + 4x^2}\right)\,dx$ we see that $\begin{array}{rcl} \displaystyle\int_0^N \left(\frac{1}{1 + 4x^2} - \cos(2x)\right) \frac{dx}{x} & = & \displaystyle \int_0^{2N} \frac{1 - \cos x}{x}\,dx - \Big[\tfrac12\ln(1 + 4x^2)\Big]_0^N \\ & = & \displaystyle \gamma + \ln(2N) - Ci(2N) - \tfrac12\ln(1 + 4N^2) \\ & = & \gamma + \ln\left(\frac{2N}{\sqrt{1+4N^2}}\right) - Ci(2N) \end{array}$ It is now clear that $\int_0^\infty \left(\frac{1}{1 + 4x^2} - \cos(2x)\right) \frac{dx}{x} \; = \; \lim_{N \to \infty} \int_0^N \left(\frac{1}{1 + 4x^2} - \cos(2x)\right) \frac{dx}{x} \; = \; \gamma \;.$

I have written another elementary solution.

I've always seen that u just provide the answer. You never provide the complete solution. Can't u just post the complete solution whenever u r willing to post an answer?