Using hypergeometric functions to solve this integral After looking at calculations, I realized that the exponent needs to be 1/4 instead of -1/4
I have this equation and I am trying to solve the integral of it. 
$$((R^2) - (y^2))^{1/4} dy$$
I tried to put it into wolfram alpha, and I got an answer, but I wanted to know how they arrived at the answer. 
Any advice would be greatly appreciated. If you could please show me how to do this integral, I would be appreciate it very much. 
I know you need to use hyper geometric functions; however, I am not sure how.
Thank-you very much 

I'm assuming the integral in question is
$$\int_0^R (R^2-y^2)^{-\frac{1}{4}}\,dy$$
as these types are quite common. If not, a very modest change can be made (just let the upper bound be $x$ instead of $R$ in the following analysis and leave the hypergeometric function in terms of $x$ and ignore the stuff concerning Gauss' theorem). Any time you come across an integral of something like $(x+y)^{\alpha}$, you should think about using the binomial theorem. The binomial theorem states that
$$ (x+y)^{\alpha} = \sum_{n=0}^{\infty} \frac{\Gamma(\alpha+1)}{\Gamma(\alpha-n+1)n!} x^{\alpha-n} y^n.$$
In our case, we get
$$ (R^2-y^2)^{-\frac{1}{4}} = \sum_{n=0}^{\infty} \frac{(-1)^n\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{3}{4}-n\right) n!} R^{-\frac{1}{2}-2n}y^{2n}.$$
Therefore
$$\int_0^R (R^2-y^2)^{-\frac{1}{4}}\,dy = \sum_{n=0}^{\infty} \frac{(-1)^n\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{3}{4}-n\right) n!}R^{-\frac{1}{2} -2n}\int_0^R y^{2n}\,dy.$$
We can interchange integral and sum via uniform convergence arguments (and considering the integral from $[0,R']$ where $R' < R$, then letting $R'\to R$). The integral gives nothing more than $\frac{1}{2n+1} R^{2n+1}$ so we are left with
$$ \int_0^R (R^2-y^2)^{-\frac{1}{4}}\,dy = R^{\frac{1}{2}} \sum_{n=0}^{\infty} \frac{(-1)^n\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{3}{4}-n\right) n!(2n+1)}. $$
This series is more than a little daunting but it can be conveniently written as $_2F_1\left(\frac{1}{4},\frac{1}{2};\frac{3}{2};1\right).$ Strictly speaking, $_2F_1$ does not exist at $1$ as the series diverges, but by considering the limit from the left, we can assign it a value at $1$. (The value could be infinity.)
Gauss' theorem states that for $_2F_1(a,b;c,z)$, the value at $1$ exists and is finite if $\Re c > \Re(a+b)$ and the value is
$$_2F_1(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}.$$
In our case, $a = \frac{1}{4}$, $b= \frac{1}{2}$ and $c=\frac{3}{2}$ so the conditions for Gauss' theorem are met and we get that the value of the series is
$$\frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{3}{2}-\frac{1}{4}-\frac{1}{2}\right)}{\Gamma\left(\frac{3}{2}-\frac{1}{4}\right)\Gamma\left(\frac{3}{2}-\frac{1}{2}\right)} = \frac{\frac{1}{2}\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)}. = \frac{\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)}{2\frac{1}{4}\Gamma\left(\frac{1}{4}\right)} = \frac{2\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}.$$
Substituting this into $(1)$, we get
$$ \int_0^R (R^2-y^2)^{-\frac{1}{4}}\,dy = \frac{2\sqrt{\pi R}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}.$$
