We're evaluating
$$I_k := \int_1^\infty \frac{dx}{1 + x^{2 k}}.$$
First, consider the case that $k$ is an integer (a positive one, which is necessary for convergence).
Mark Viola's answer to a related question gives that the indefinite integral is given by
\begin{multline}
I_k(x)
:= \int\frac{dx}{1 + x^{2 k}} \\
= \sum_{n=1}^{2 k}\underbrace{-\frac1{2 k}\left(\frac12 \cos \alpha_n \log(x^2-2\cos \alpha_n \cdot x+1)-\sin \alpha_n \arctan\left(\frac{x-\cos \alpha_n}{\sin \alpha_n}\right)\right)}_{A_n(x)}+C ,
\end{multline}
where for compactness we denote $$\alpha_n := \frac{(2 n - 1) \pi}{2 k} .$$
The Fundamental Theorem of Calculus gives that our definite integral is
$$I_k = \lim_{x \to \infty} (I_k(x) - I_k(1)).$$ Now, the individual summands $A_n(x)$ diverge as $x \to \infty$, but owing to the fact that $2 k$ is even, in the "antipodal" sums $A_m(x) + A_{k + m}(x)$ the logarithmic terms cancel as $x \to \infty$, so the limits $\lim_{x \to \infty} (A_m(x) + A_{k + m}(x))$ converge, and our integral is given by
$$I_k = \sum_{m = 1}^k \left[A_m(x) + A_{k + m}(x)\right]_{x = 1}^\infty .$$
A straightforward computation evaluation gives
$$\lim_{x \to \infty} (A_m(x) + A_{m + k}(x)) = -\frac\pi{2k} \sin \frac{(2 \pi - 1) n}{2 k},$$ and summing over $m$ gives
$$\lim_{x \to \infty} I_k(x) = \sum_{m = 1}^k (A_m(x) + A_{m + k}(x)) = \sum_{m = 1}^k -\frac\pi{2k} \sin \frac{(2 \pi - 1) n}{2 k} = \frac{\pi \sin \frac\pi{2 k}}{k \left(1 - \cos \frac\pi k\right)} .$$
On the other hand, substituting gives
\begin{align}
I_k(1)
&= -\frac1{2k} \sum_{n = 1}^{2 k} \left(\frac12 \cos \alpha_n \log (2 - 2 \cos \alpha_n) - \sin \alpha_n \arctan \left(\frac{1 - \cos \alpha_n}{\sin \alpha_n}\right)\right) \\
&= -\frac1{2k} \sum_{n = 1}^{2 k} \left(\frac12 \cos \alpha_n \log (2 - 2 \cos \alpha_n) - \frac{(2 n - 1) \pi}{4 k} \sin \alpha_n \right) ;
\end{align}
here the second equality follows from the tangent half-angle identity: $\frac{1 - \cos \alpha_n}{\sin \alpha_n} = \tan \frac{\alpha_n}2$.
So, our original integral is
$$\small \boxed{I_k = \frac{\pi \sin \frac\pi{2 k}}{k \left(1 - \cos \frac\pi k\right)} + \frac1{2 k} \sum_{n = 1}^{2 k} \left(\frac12 \cos \frac{(2 n - 1) \pi}{2 k} \log \left(2 - 2 \cos \frac{(2 n - 1) \pi}{2 k}\right) - \frac{(2 n - 1) \pi}{4 k} \sin \frac{(2 n - 1) \pi}{2 k}\right)} . $$
Perhaps this expression can be simplified further still in terms of elementary functions.
The first several values are
$$I_1 = \frac\pi4, \qquad
I_2 = \frac1{2 \sqrt 2} (\pi - \operatorname{arcosh} 3), \qquad
I_3 = \frac16 (\pi - \sqrt 3 \operatorname{arcosh} 2) .$$
We can extend our results to arbitrary, i.e., not necessarily integral, exponent $k > \frac12$ (that restriction is forced by convergence). Notice that if $k = \frac pq$ is rational ($p, q > 0$), then the substitution $x = u^q$ rationalizes the integrand, giving $q \int_1^\infty \frac{u^{q - 1} \,du}{1 + u^{2 p}}$ so in that case $I_k$ still admits a closed-form expression in terms of elementary functions. For typical irrational values, however, that will not be the case.
Digamma function Rewrite the integrand as a series in $\frac1 x$:
$$\frac1{1 + x^{2k}} = \frac1{x^{2k}} \frac1{1 + x^{-2k}} = x^{-2k}\sum_{i=0}^\infty(-1)^i x^{-2 k i} = \sum_{i=0}^\infty(-1)^{i} x^{-2 k (1 + i)}.$$
Integrating gives
\begin{multline*}
\int_1^\infty \frac{dx}{1 + x^{2k}}
= \int_1^\infty \sum_{n=0}^\infty(-1)^n x^{-2 k (1 + n)} \,dx
= \sum_{i=0}^\infty(-1)^n \int_1^\infty x^{-2 k (1 + n)} \,dx
\\ = \sum_{n=0}^\infty \frac{(-1)^n}{(2 k + 1) n - 1}
= \boxed{\frac1{4k} \left[\psi \left(1 - \frac1{4k}\right) - \psi\left(\frac12 - \frac1{4k}\right)\right]} ,
\end{multline*}
where $\psi := \frac{\Gamma'}{\Gamma}$ is the digamma function. The last equality follows from the identity
$$\psi(z) = -\gamma + \sum_{n=0}^\infty \left(\frac{1}{1 + n} - \frac1{z + n}\right) ;$$
here $\gamma$ is the Euler–Mascheroni constant. This approach has the advantage that the usual asymptotic series for the digamma function immediately yields an asymptotic formula for $I_k$, namely,
$$I_k \sim \frac{\log 2}{2 k} + \frac{\pi^2}{48 k^2} + \frac{3 \zeta(3)}{32 k^3} + \cdots,$$
where $\zeta(3)$ is Apéry's constant and $\cdots$ denotes a remainder in $O(k^{-4})$.
Incomplete beta function Instead substituting $x = \cot^{\frac1k} \theta$ transforms the integral to
$$\frac1k \int_0^\frac\pi4 \sin^{1-\frac1k} \cos^{\frac1k-1} \theta \,d\theta = \boxed{\operatorname{B}\left(\frac12; 1 - \frac1{2k}, \frac1{2k}\right)} ,$$ where $\operatorname B$ now denotes the incomplete beta function.
Ordinary hypergeometric function The incomplete beta function is a special case of the ordinary hypergeometric function, ${}_2 F_1$, and so we can write the value of the integral, for example, as
$$\frac1{2 k - 1}\cdot{}_2 F_1 \left(1, 1 - \frac1{2k}; 2 -\frac1{2k}; -1\right) .$$
