<s>[INST] Evaluating $\int_0^1 \sqrt{1 + x ^4 } \, d x $ $$
\int_{0}^{1}\sqrt{\,1 + x^{4}\,}\,\,\mathrm{d}x
$$ 
I used substitution of tanx=z but it was not fruitful. Then i used $ (x-1/x)= z$ and
$(x)^2-1/(x)^2=z $ but no helpful expression was derived.
I also used property $\int_0^a f(a-x)=\int_0^a f(x) $
Please help me out
 [/INST] Consider the ${}_{2}F_{1}$ hypergeometric integral form given by
$${}_{2}F_{1}(a, b; c; x) = \frac{\Gamma(c)}{\Gamma(b) \, \Gamma(c-b)} \, \int_{0}^{1} t^{b-1} \, (1-t)^{c-b-1} \, (1-x \, t)^{-a} \, dt$$
leads to, with $a=-1/2$, $b=1/4$, $c=5/4$, $x=-1$,
$${}_{2}F_{1}\left(-\frac{1}{2}, \frac{1}{4}; \frac{5}{4}; -1\right) = \frac{1}{4} \, \int_{0}^{1} t^{-3/4} \, \sqrt{1+ t} \, dt.$$
Now let $t = x^{4}$ to obtain
$$\int_{0}^{1} \sqrt{1 + x^4} \, dx = {}_{2}F_{1}\left(-\frac{1}{2}, \frac{1}{4}; \frac{5}{4}; -1\right) = 1.08943...$$
</s>