What is the the integral of $\sqrt{x^a + b}$? How do you evaluate $\displaystyle\int\sqrt{x^a + b}\,\,\text{dx}$, where $a \neq 0$ and $a \neq 1$?
For example, how do you evaluate $\displaystyle\int\sqrt{x^2 + 1}\,\text{dx}$? If we let $u=x^2+1$, then $du=2x\,\text{dx}$. We cannot do this because there is no $2x$ in the original function. Of course you cannot let $u=\sqrt{x^2 + 1}$, because $du=\displaystyle\frac{x}{\sqrt{x^2+1}}\,\text{dx}$. There is no $\displaystyle\frac{x}{\sqrt{x^2+1}}$ in the original function. So how do we solve this?
How about $\displaystyle\int\sqrt{x^3 + 1}\,\text{dx}$?

Kim Peek's "funny" hypergeometric solution is really  the series solution near $x=0$.  We have, for $|x^a/b| < 1$,
$$ \sqrt{x^a + b} = \sqrt{b} \sqrt{1 + x^a/b} = 
\sqrt{b} \sum_{k=0}^\infty {1/2 \choose k} (x^a/b)^k$$
so integrating term-by-term
$$ \int \sqrt{x^a + b}\; dx =  \sum_{k=0}^\infty {1/2 \choose k} \dfrac{x^{ak+1}}{(ak+1) \; b^{k-1/2}}$$
where $${1/2 \choose k} = \dfrac{\Gamma(3/2)}{\Gamma(k+1)\; \Gamma(3/2-k)}
= \dfrac{(2k)!}{(-4)^k (k!)^2 (1-2k)}$$
If you take the definition of the hypergeometric function as a power series, you get exactly this series.
Alternatively, for $|x^a/b| > 1$ you get a different series involving negative powers of $x$:
$$\sqrt{x^a + b} = x^{a/2} \sqrt{1 + b/x^a} = \sum_{k=0}^\infty {1/2 \choose k} b^k x^{(1/2 - k)a}$$
so that
$$ \int \sqrt{x^a + b}\; dx = \sum_{k=0}^\infty {1/2 \choose k} \dfrac{b^k x^{(1/2-k)a+1}}{(1/2-k)a + 1}$$
which also has a hypergeometric representation.
