integral $\int_0^1\frac{1}{\sqrt{1+x^4}}\text{d}x$ 
How to prove that $$0.78<\int_0^1\frac{1}{\sqrt{1+x^4}}\text{d}x<0.93$$

approch: $x^2<\sqrt{1+x^4}<x^2+\frac{1}{2x^2}$
Any hint would be appreciated.

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*}
