An Arctan integral via a Quartic equation.

A series of variable changes ( $x = u^4$ , $t = s^4$ , $v = su$ ) converts the integral as follows: $\begin{aligned} \int_0^\infty \frac{\tan^{-1}x}{1 + x} \, \frac{dx}{x^{\frac14}} & = \; \int_0^\infty \frac{1}{(1+x)x^{\frac14}} \int_0^1 \frac{x\,dt}{1 + x^2t^2}\,dx \; =\; \int_0^1\left(\int_0^\infty \frac{x^{\frac34}\,dx}{(1 + x)((1 + x^2t^2)}\right)\,dt \\ & = \; 4\int_0^1 \left(\int_0^\infty \frac{u^3 \times u^3\,du}{(1 + u^4)(1 + t^2u^8)}\right)\,dt \; = \; 2\int_0^1 \left(\int_\mathbb{R} \frac{u^6\,du}{(1 + u^4)(1 + t^2u^8)}\right)\,dt \\ & = \; 8\int_0^1 \left(\int_\mathbb{R} \frac{u^6\,du}{(1 + u^4)(1 + s^8u^8)}\right)\,s^3\,ds \; = \; 8\int_0^1 \left(\int_\mathbb{R} \frac{v^6\,dv}{(v^4 + s^4)(v^8 + 1)}\right)\,ds \end{aligned}$ The integral $\int_\mathbb{R}\frac{v^6\,dv}{(v^4 + s^4)(v^8 + 1)}$ can be evaluated by standard contour integration tricks, adding up six residues in the upper half plane, and we obtain that it equals (assuming $0 < s < 1$ ) $\frac{\pi}{\sqrt{2}(1 + s^8)}\left\{ \frac{\sqrt{2-\sqrt{2}}}{2} - s^3 + (1 + \sqrt{2})\frac{\sqrt{2-\sqrt{2}}}{2}s^4\right\}$ To evaluate the original integral, we just need to evaluate the three integrals $\int_0^1 \frac{ds}{1 + s^8} \hspace{2cm} \int_0^1 \frac{s^3\,ds}{1 + s^8} \hspace{2cm} \int_0^1 \frac{s^4\,ds}{1 + s^8}$ The second of these is equal to $\tfrac{1}{16}\pi$ , and the other two can be evaluated by some involved partial fractions calculations. The end result is that $\int_0^\infty \frac{\tan^{-1}x}{1 + x} \, \frac{dx}{x^{\frac14}} \; =\; \frac{\pi}{\sqrt{2}}\left\{ \frac{\pi}{2} + 2\mathrm{arccoth}\,\sqrt{2(2+\sqrt{2})} \right\} \; = \; \frac{\pi}{\sqrt{2}}\left\{ \frac{\pi}{2} + \ln\left(\frac{\sqrt{2(2+\sqrt{2})}+1}{\sqrt{2(2+\sqrt{2})}-1}\right) \right\}$ so that $\beta \; = \; \frac{\sqrt{2(2+\sqrt{2})}+1}{\sqrt{2(2+\sqrt{2})}-1}$ It is easy to check that $\beta$ has minimal polynomial $X^4 - 28X^3 + 70X^2 - 28X + 1$ making the answer to the question $\boxed{-1}$ .