Using series find $\int_0^1 \sqrt{1+x^4}\hspace{1mm} dx$ up to $2$ decimal places I cannot figure out an aesthetic way to do this.
Can someone give a beautiful solution to this ugly question?

This is what I have tried yet. 
I used the fact that $$x = \dfrac{1}{1-\left(1-\dfrac{1}{x}\right)}$$
Hence $$x = \sum_{n=0}^{\infty} \left(1-\dfrac{1}{x}\right)^n$$
Now replacing $x$ with $\sqrt{1+x^4}$, to get $$\sqrt{1+x^4} = \sum_{n=0}^{\infty} \left(1-\dfrac{1}{\sqrt{1+x^4}}\right)^n$$
$ $
Just wondering if there are better alternatives available.
 A: Hint :
Maclaurin series for Binomial series is
$$
(1+y)^n=\sum_{k=0}^\infty\binom{n}{k} y^n,
$$
for $|y|<1$ and all $n$. Also use identities  binomial coefficients:
$$
\binom{n}{0}=1
$$
and
$$
\binom{n}{k+1}=\binom{n}{k}\frac{n-k}{k+1}.
$$
Now, replace $y=x^4$ and then integrate it term by term.
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\int_{0}^{1}\root{1 + x^{4}}\,\dd x&=
\int_{0}^{1}\sum_{n = 0}^{\infty}{1/2 \choose n}x^{4n}\,\dd x
=\sum_{n = 0}^{\infty}{1/2 \choose n}\int_{0}^{1}x^{4n}\,\dd x
=\sum_{n = 0}^{\infty}{1/2 \choose n}{1 \over 4n + 1}
\end{align}

Now we 'cut' the series once the absolute value of the sum 'general term'
  $\ds{{1/2 \choose n}{1 \over 4n + 1}}$ becomes smaller than $\ds{10^{-2}}$: 
  $$
\begin{array}{rclrcl}
\verts{{1/2 \choose n}{1 \over 4n + 1}}_{\,n\ =\ 0} & = & 1\,,&\qquad
\verts{{1/2 \choose n}{1 \over 4n + 1}}_{\,n\ =\ 1} & = & 0.1
\\[3mm]
\verts{{1/2 \choose n}{1 \over 4n + 1}}_{\,n\ =\ 2} & \approx &-0.0139\,,&\qquad
\verts{{1/2 \choose n}{1 \over 4n + 1}}_{\,n\ =\ 3} & \approx & 0.0048
\end{array}
$$

Then, I'll use three terms $\ds{\pars{~n = 0,1,2}}$:
\begin{align}
\color{#66f}{\large\int_{0}^{1}\root{1 + x^{4}}\,\dd x}
&\approx {1/2 \choose 0}\,{1 \over 4\times 0 + 1}
+{1/2 \choose 1}\,{1 \over 4\times 1 + 1} + {1/2 \choose 2}\,{1 \over 4\times 2 + 1}
\\[3mm]&=1 + 0.1 - 0.0139 = \color{#66f}{\large 1.08\color{#c00000}{61}} 
\end{align}

A numerical calculation of the integral yields
  $\ds{\approx \color{#66f}{\large 1.08}9429413}$.

