integral $\int_0^1\frac{1}{\sqrt{1+x^4}}\text{d}x$

This is so fun :) The $>$ part is easy because we know that for $0<x<1$, $1+x^4<1+x^2$, so $$\int_0^1\frac1{\sqrt{1+x^4}}dx>\int_0^1\frac1{\sqrt{1+x^2}}dx$$ Let $x=\sinh\theta$. Then $$\int_0^1\frac1{\sqrt{1+x^4}}dx>\int_0^{\sinh^{-1}(1)}d\theta=\sinh^{-1}(1)\approx0.881374$$ The $<$ part I found more difficult. If we use the Taylor series to estimate $(1+x^4)^{-1/2}$, we have $$\begin{align}\frac1{\sqrt{1+x^4}}-\left(1-\frac12x^4\right) & =\frac{1-\sqrt{1+x^4}\left(1-\frac12x^4\right)}{\sqrt{1+x^4}}\\ & =\frac{1-\left(1+x^4\right)\left(1-x^4+\frac14x^2\right)}{\sqrt{1+x^4}\left(1+\sqrt{1+x^4}\left(1-\frac12x^4\right)\right)}\\ & =\frac{\frac34x^8-\frac14x^{12}}{\sqrt{1+x^4}+1+\frac12x^4-\frac12x^8}\\ & <\frac{\frac34x^8}{2}=\frac38x^8\end{align}$$ So now $$\int_0^1\frac1{\sqrt{1+x^4}}dx<\int_0^1\left(1-\frac12x^4+\frac38x^8\right)dx=1-\frac1{10}+\frac1{24}\approx0.941667$$ Oops, just noticed that fixing an arithmetic error along the way spoiled my estimate. Have to recalibrate.

EDIT: OK, the bad news is that it takes 9 nonzero terms of the Taylor series to get a high sum $<0.93$. The good news is that the taylor series is alternating so we know that it is for sure an overestimate. For $0<x<1$, $$\frac1{\sqrt{1+x^4}}<1-\frac12x^4+\frac38x^8-\frac5{16}x^{12}+\frac{35}{128}x^{16}-\frac{63}{256}x^{20}+\frac{231}{1024}x^{24}-\frac{429}{2048}x^{28}+\frac{6435}{32768}x^{32}$$ So $$\int_0^1\frac1{\sqrt{1+x^4}}dx<1-\frac12\left(\frac1{5}\right)+\frac38\left(\frac1{9}\right)-\frac5{16}\left(\frac1{13}\right)+\frac{35}{128}\left(\frac1{17}\right)-\frac{63}{256}\left(\frac1{21}\right)+\frac{231}{1024}\left(\frac1{25}\right)-\frac{429}{2048}\left(\frac1{29}\right)+\frac{6435}{32768}\left(\frac1{33}\right)\approx0.929745$$ The upper bound is so tight that it's hard to get a good enough estimate.

I know this doesn't constitute as a proof, but I thought the following was interesting. It turns out we can use the Beta function to express your integral in terms of the Gamma function by taking advantage of the fact that

\begin{align*} B(x,y) = \int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}}dt. \end{align*}

First, substituting $x=t^4$, we see that

\begin{align*} \int_0^1\frac{dx}{\sqrt{1+x^4}}= \frac{1}{4}\int_0^1 \frac{t^{-3/4}}{(t+1)^{1/2}} \, dt. \end{align*} By making a substitution such as $u=1/t$, after simplification one ends up with an identical integral but instead over $(1,\infty)$, so we can deduce that \begin{align*} B(1/4,1/4)=\int_0^\infty \frac{t^{-3/4}}{(t+1)^{1/2}} \, dt=2\int_0^1 \frac{t^{-3/4}}{(t+1)^{1/2}} \, dt. \end{align*} It follows that \begin{align*} \frac{1}{4}\int_0^1 \frac{t^{-3/4}}{(t+1)^{1/2}}dt &= \frac{B(1/4,1/4)}{8}\\ &= \frac{\Gamma(1/4)\Gamma(1/4)}{8\Gamma(1/2)}\\ &=\frac{\Gamma(1/4)^2}{8\sqrt \pi}. \end{align*}

From here there are probably ways of getting algebraic bounds that would be better for a proof. This particular value of Gamma has been approximated to death however, and so we see that your original integral is \begin{align*} \int_0^1\frac{dx}{\sqrt{1+x^4}} = \frac{\Gamma(1/4)^2}{8\sqrt \pi} \approx 0.927037. \end{align*}

Even though my previous answer solved the problem I hated it because there is no way when the problem was posed that it could have been intended to require $9$ terms of a slowly converging Taylor series. If you let $x^2=\tan(\frac{\theta}2)$ then $$\int_0^1\frac1{\sqrt{1+x^4}}dx=\frac{\sqrt2}4\int_0^{\frac{\pi}2}\frac{d\theta}{\sqrt{\sin\theta}}=\frac{\sqrt2}8\text{B}\left(\frac14,\frac12\right)=\frac{\left(\Gamma\left(\frac14\right)\right)^2}{8\sqrt{\pi}}\approx0.927037$$ Since this is so close to the desired upper bound, a really good estimate must be made. Just recently I showed that $$\begin{align}\int\frac{du}{u^4+1}= & -\frac1{4\sqrt2}\ln\left(u^2-\sqrt2u+1\right)+\frac1{2\sqrt2}\tan^{-1}\left(\sqrt2u+1\right)+\\ &\frac1{4\sqrt2}\ln\left(u^2+\sqrt2u+1\right)+\frac1{2\sqrt2}\tan^{-1}\left(\sqrt2u-1\right)+C\\ &=f(u)+C\end{align}$$ We can leverage this result to get estimates for the integral in question as follows: $$\begin{align}\int_0^1\frac1{\sqrt{1+x^4}}dx & >\int_0^1\frac1{\sqrt{1+x^4+\frac14x^8}}dx=\int_0^1\frac1{1+\frac12x^4}dx\\ & =2^{\frac14}\int_0^{\left(\frac12\right)^{\frac14}}\frac{du}{u^4+1}=2^{\frac14}\left(f\left(\left(\frac12\right)^{\frac14}\right)-f(0)\right)\\ & \approx0.920788\end{align}$$ We have the good estimate $$\frac1{\sqrt{1+x^4}}-1=\frac{1-\sqrt{1+x^4}}{\sqrt{1+x^4}}=\frac{-x^4}{\sqrt{1+x^4}+1+x^4}$$ So $$\begin{align}\int_0^1\frac1{\sqrt{1+x^4}}dx & =\int_0^1\left(1-\frac{x^4}{\sqrt{1+x^4}+1+x^4}\right)dx\\ & <1-\int_0^1\frac{x^4}{\sqrt{1+x^4+\frac14x^8}+1+x^4}dx\\ & =1-\int_0^1\frac{x^4}{2+\frac32x^4}dx=1-\frac12\left(\frac43\right)^{\frac54}\int_0^{\left(\frac34\right)^{\frac14}}\frac{u^4}{1+u^4}du\\ & =1-\frac12\left(\frac43\right)^{\frac54}\left(\left(\frac34\right)^{\frac14}-f\left(\left(\frac34\right)^{\frac14}\right)+f(0)\right)\\ & \approx0.927798\end{align}$$ I think this approach is more consistent with the spirit of the problem.

EDIT: There is also a way of estimating that integral in the upper bound without resorting to that intimidating interal used above. Since $$\frac{x^4}{\sqrt{1+x^4}+1+x^4}\ge0$$ and $$0\le\frac18(15x-10x^3+3x^5)\le1$$ for $0\le x\le1$, we can make the substitution $x^2=\sinh\theta$ to show that $$\begin{align}\int_0^1\frac{x^4}{\sqrt{1+x^4}+1+x^4}dx & >\frac18\int_0^1(15x-10x^3+3x^5)\frac{x^4}{\sqrt{1+x^4}+1+x^4}dx\\ & =\frac18\int_0^{\sinh^{-1}(1)}\frac{\left(15-10\sinh\theta+3\sinh^2\theta\right)\sinh^2\theta}{\cosh\theta+\cosh^2\theta}\cdot\frac12\cosh\theta\,d\theta\\ & =\frac1{16}\int_0^{\sinh^{-1}(1)}\frac{\left(15-10\sinh\theta+3\sinh^2\theta\right)\left(\cosh^2\theta-1\right)}{1+\cosh\theta}\,d\theta\\ & =\frac1{16}\int_0^{\sinh^{-1}(1)}\left(15-10\sinh\theta+3\sinh^2\theta\right)\left(\cosh\theta-1\right)\,d\theta\\ & =\frac1{16}\left[15\sinh\theta-5\sinh^2\theta+\sinh^3\theta-15\theta+10\cosh\theta-\frac32\left(\frac12\sinh2\theta-\theta\right)\right]_0^{\sinh^{-1}(1)}\\ & =\frac1{16}\left[15-5+1-15\sinh^{-1}(1)+10\left(\sqrt2-1\right)-\frac32\left(\sqrt2-\sinh^{-1}(1)\right)\right]\\ & =\frac1{16}\left[1-\frac{27}2\ln\left(1+\sqrt2\right)+17/2\sqrt2\right]\\ & \approx0.070142\end{align}$$ So then $$\begin{align}\int_0^1\frac1{\sqrt{1+x^4}}dx & =1-\int_0^1\frac{x^4}{\sqrt{1+x^4}+1+x^4}dx\\ & <1-\frac1{16}\left[1-\frac{27}2\ln\left(1+\sqrt2\right)+17/2\sqrt2\right]\approx0.929858\end{align}$$ Not as good an estimate, but (just barely) good enough, and a much easier integral.