<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] 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.
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