Estimate $\int_0^{\infty} 1/\sqrt{1+x^4} \mathrm{d}x$
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \int_{0}^{\infty}{\dd x \over \root{1 + x^{4}}}&= {1 \over 4}\int_{0}^{\infty}x^{-3/4}\pars{1 + x}^{-1/2}\,\dd x = {1 \over 4}\int_{1}^{\infty}\pars{x - 1}^{-3/4}x^{-1/2}\,\dd x \\[3mm]&= {1 \over 4}\int_{1}^{0}\pars{{1 \over x} - 1}^{-3/4}x^{1/2}\, \pars{-\,{\dd x \over x^{2}}} = {1 \over 4}\int^{1}_{0}\pars{1 - x}^{-3/4}x^{-3/4}\,\dd x \\[3mm]&= {1 \over 4}\,{\rm B}\pars{{1 \over 4},{1 \over 4}} \quad\mbox{where}\quad{\rm B}\pars{a,b}\quad\mbox{is the}\ {\it Beta\ function}. \end{align} Also $$ {\rm B}\pars{{1 \over 4},{1 \over 4}} = {\Gamma\pars{1/4}\Gamma\pars{1/4} \over \Gamma\pars{1/4 + 1/4}} ={1 \over \root{\pi}}\,\Gamma^{2}\pars{1 \over 4} $$ $$\color{#00f}{\large% \int_{0}^{\infty}{\dd x \over \root{1 + x^{4}}} = {1 \over 4\root{\pi}}\,\Gamma^{2}\pars{1 \over 4}} \approx 1.85 $$
Actually, to educate the OP a little - you can use complex variables techniques here. You just have to avoid the branch points. Consider the following integral:
$$\oint_C \frac{dz}{\sqrt{1+z^4}} $$
where $C$ is the following contour:
We can then write out the contour integral explicitly in terms of a parametrization; the various terms are
$$\int_{-R}^R \frac{dx}{\sqrt{1+x^4}} + i R \int_0^{\pi} d\theta \, \frac{e^{i \theta}}{\sqrt{1+R^4 e^{i 4 \theta}}} \\ + e^{i \pi/4} \int_1^{R}dt \frac{e^{i \pi/2}}{\sqrt{t^4-1}}- e^{i \pi/4} \int_1^{R}dt \frac{e^{-i \pi/2}}{\sqrt{t^4-1}}\\ + i \epsilon \int_{2 \pi}^0 d\phi \, \frac{e^{i \phi}}{\sqrt{1+(e^{i \pi/4}+\epsilon e^{i \phi})^4}} \\ + e^{i 3\pi/4} \int_1^{R}dt \frac{e^{i \pi/2}}{\sqrt{t^4-1}}- e^{i 3\pi/4} \int_1^{R}dt \frac{e^{-i \pi/2}}{\sqrt{t^4-1}}\\ + i \epsilon \int_{2 \pi}^0 d\phi \, \frac{e^{i \phi}}{\sqrt{1+(e^{i 3\pi/4}+\epsilon e^{i \phi})^4}} $$
Note that the factors of $e^{i \pi/2}$ and $e^{-i \pi/2}$ are a result of the $2 \pi$ jumps about the branch points.
In the limit as $R \to \infty$ and $\epsilon \to 0$, the second, fifth, and eighth integrals vanish. Simplifying, we are left with
$$\int_{-\infty}^{\infty} \frac{dx}{\sqrt{1+x^4}} + i 2 \left ( e^{i \pi/4} + e^{i 3 \pi/4}\right ) \int_1^{\infty} \frac{dt}{\sqrt{t^4-1}}$$
Noting that the contour integral is zero by Cauchy's theorem, we may finally deduce that
$$\int_0^{\infty} \frac{dx}{\sqrt{1+x^4}} = \sqrt{2} \int_1^{\infty} \frac{dt}{\sqrt{t^4-1}}$$
The integral on the RHS may be simplified by subbing $t=1/y$ to get
$$\sqrt{2} \int_0^1 \frac{dy}{\sqrt{1-y^4}}$$
Then sub $y = u^{1/4}$ to get
$$\frac{\sqrt{2}}{4} \int_0^1 du \, u^{-3/4} (1-u)^{-1/2}$$
which is a Beta function. Thus the integral is
$$\frac{\sqrt{2}}{4} \frac{\Gamma \left ( \frac14\right ) \Gamma \left ( \frac12\right )}{\Gamma \left ( \frac{3}{4}\right )} = \frac{4}{\sqrt{\pi}} \Gamma \left ( \frac{5}{4}\right )^2$$
The RHS may be derived using the reflection formula.
For $x \in [0,\infty]$, let $I(x)$ be the integral $\;\displaystyle\int_0^x \frac{dt}{\sqrt{1+t^4}}$.
The integral we want to calculate is simply $I(\infty)$. Quoting some results from this answer of a related question, we know for $x \in [0,\infty)$,
$$I(x) = F(\sqrt{1+x^2-\sqrt{1+x^4}}; \frac12 )$$ where $$F(y\,; m) = \int_0^y \frac{ds}{\sqrt{(1-s^2)(1-ms^2)}}$$ is the Jacobi's form of incomplete elliptic integral of the first kind.
When $x \to \infty$, $\sqrt{1 + x^2 - \sqrt{1+x^4}} \to 1$. This leads to
$$I(\infty) = K\left(\frac12\right)\quad\text{ where }\quad K(m) = \int_0^1 \frac{ds}{\sqrt{(1-s^2)(1-ms^2)}}$$ is the complete elliptic integral of the first kind. $K(m)$ can be computed efficiently using its relation with the arithmetic geometric mean of $1$ and $\sqrt{1-m}$.
$$K(m) = \frac{\pi}{2\text{AGM}(1,\sqrt{1-m})}$$
Start with $m = \frac12$, the first iteration of computing the AGM gives us
$$ \frac{1}{\sqrt[4]{2}} = \text{GM}(1,\frac{1}{\sqrt{2}} ) \le \text{AGM}(1,\frac{1}{\sqrt{2}} ) \le \text{AM}(1, \frac{1}{\sqrt{2}} ) = \frac12 (1 + \frac{1}{\sqrt{2}})$$ and hence $$ 1.840302369 \sim \frac{\sqrt{2}\pi}{\sqrt{2}+1} \le I(\infty) = K\left(\frac12\right) \le \frac{\pi}{2^{3/4}} \sim 1.868002168 $$
To get more accurate estimate of the integral, one can carry out more iterations in the AGM computation. For example, the second iteration gives us
$$ 1.854048814 \sim \frac{2^{3/2}\pi }{2^{5/4}+2^{1/2}+1} \le K\left(\frac12\right) \le \frac{\pi }{\sqrt{2^{1/4} + 2^{3/4}}} \sim 1.8541005407$$ which is accurate to about $4^{th}$ decimal places.