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
 A: 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
$$\int_{0}^{1} \sqrt{1 + x^4} \, dx = {}_{2}F_{1}\left(-\frac{1}{2}, \frac{1}{4}; \frac{5}{4}; -1\right) = 1.08943...$$
A: We can do better than elliptic integrals or $\Gamma$ / Beta / hypergeometric functions:
$$ \int_{0}^{1}\sqrt{1+t^4}\,dt = \frac{\sqrt{2}}{3}+\frac{\pi}{6\,\text{AGM}\left(1,\frac{1}{\sqrt{2}}\right)} \tag{1}$$
due to integration by parts and what is shown in this answer. $(1)$ summarizes a very efficient numerical technique (the $\text{AGM}$ iteration has a quadratic convergence) for computing arbitrarily accurate numerical approximations of the LHS. It also shows
$$ 1.084\ldots=\frac{\sqrt{2}}{3}+\frac{\pi}{3}(2-\sqrt{2})\leq \int_{0}^{1}\sqrt{1+t^4}\,dt\leq \frac{\sqrt{2}}{3}+\frac{\pi}{6}2^{1/4}=1.094\ldots\tag{2}$$
Improving the bound $\leq\sqrt{\frac{6}{5}}$ given by Jensen's inequality. Your integral is strictly related to the lemniscate constant.
A: Another non-elementary answer, from Maple, is 
$$ \int_0^1 \sqrt{1+x^4}\; dx = \frac{\sqrt {2}+{\it EllipticK} \left( 1/\sqrt {2} \right)}{3} $$
A: We can do better than hypergeometric function and elliptic integral:
$$\color{blue}{\int_0^1 {\sqrt {1 + {x^4}} dx}  = \frac{{\sqrt 2 }}{3} + \frac{{{\Gamma ^2}(\frac{1}{4})}}{{12\sqrt \pi  }}}$$

Firstly, integration by part gives
$$\int_0^1 {\sqrt {1 + {x^4}} dx}  = \sqrt 2  - 2\int_0^1 {\frac{{{x^4}}}{{\sqrt {1 + {x^4}} }}dx}  = \sqrt 2  - 2\int_0^1 {\left( {\sqrt {1 + {x^4}}  - \frac{1}{{\sqrt {1 + {x^4}} }}} \right)dx} $$
Hence
$$\int_0^1 {\sqrt {1 + {x^4}} dx}  = \frac{{\sqrt 2 }}{3} + \frac{2}{3}\int_0^1 {\frac{1}{{\sqrt {1 + {x^4}} }}dx} $$
Making $x=1/u$ in the last integral gives
$$\int_0^1 {\frac{1}{{\sqrt {1 + {x^4}} }}dx}  = \frac{1}{2}\int_0^\infty  {\frac{1}{{\sqrt {1 + {x^4}} }}} dx = \frac{1}{{8\sqrt \pi  }}{\Gamma ^2}(\frac{1}{4})$$
which can be evaluated by using some formula for beta function.
A: For an approximation, you could use a Padé approximant for the integrand. The simplest one would be
$$\frac{3 x^4+4}{x^4+4}=3-\frac{x+2}{x^2+2 x+2}+\frac{x-2}{x^2-2 x+2}$$
$$\int \frac{3 x^4+4}{x^4+4}\,dx=3x+\frac{1}{2} \log \left(\frac{x^2-2 x+2}{x^2+2 x+2}\right)+\tan ^{-1}(1-x)-\tan ^{-1}(1+x)$$ So, using the given bounds, an approximation is 
$$\int_{0}^{1}\sqrt{\,1 + x^{4}\,}\,\,dx \approx 3-\frac{\log (5)}{2}-\tan ^{-1}(2)\approx 1.08813$$
