Answer
$\mathop \smallint \limits_0^x \dfrac{{{\rm{d}}t}}{{\sqrt {1 - {t^4}} }} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right)\dfrac{{{{\left( { - 1} \right)}^n}{x^{4n + 1}}}}{{4n + 1}}$
Work Step by Step
Let us expand ${\left( {1 + t} \right)^{ - 1/2}}$ by Theorem 3:
${\left( {1 + t} \right)^a} = 1 + \dfrac{a}{{1!}}t + \dfrac{{a\left( {a - 1} \right)}}{{2!}}{t^2} + \dfrac{{a\left( {a - 1} \right)\left( {a - 2} \right)}}{{3!}}{t^3} + \cdot\cdot\cdot + \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){t^n} + \cdot\cdot\cdot$
converges for $\left| t \right| \lt 1$.
${\left( {1 + t} \right)^a} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
a\\
n
\end{array}} \right){t^n}$
${\left( {1 + t} \right)^{ - 1/2}} = \mathop \sum \limits_{n = 0}^\infty\left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right){t^n}$
converges for $\left| t \right| \lt 1$.
Substituting $ - {t^4}$ for $t$ in the series above gives
${\left( {1 - {t^4}} \right)^{ - 1/2}} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right){\left( { - 1} \right)^n}{t^{4n}}$
converges for $\left| { - {t^4}} \right| \lt 1$, or $\left| t \right| \lt 1$.
Thus, $\dfrac{1}{{\sqrt {1 - {t^4}} }} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right){\left( { - 1} \right)^n}{t^{4n}}$.
Taking the definite integral on the series gives
$\mathop \smallint \limits_0^x \dfrac{{{\rm{d}}t}}{{\sqrt {1 - {t^4}} }} = \mathop \sum \limits_{n = 0}^\infty \mathop \smallint \limits_0^x \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right){\left( { - 1} \right)^n}{t^{4n}}{\rm{d}}t$
$ = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right){\left( { - 1} \right)^n}\left[ {\dfrac{{{t^{4n + 1}}}}{{4n + 1}}} \right]_0^x = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right)\dfrac{{{{\left( { - 1} \right)}^n}{x^{4n + 1}}}}{{4n + 1}}$
So, $\mathop \smallint \limits_0^x \dfrac{{{\rm{d}}t}}{{\sqrt {1 - {t^4}} }} = \mathop \sum \limits_{n = 0}^\infty \left( {\begin{array}{*{20}{c}}
{ - \dfrac{1}{2}}\\
n
\end{array}} \right)\dfrac{{{{\left( { - 1} \right)}^n}{x^{4n + 1}}}}{{4n + 1}}$.
