Numerical integration of $\int_a^bf(x) \: \text{d}x$ for $f(x) \to \infty$ when $x \to b$ This is the function I am trying to approximate using Simpson's rule: 

$$\int_0^1 f(x) \: \text{d}x =\int_0^1 \frac{e^x}{\sqrt{1-x^2}} \: \text{d}x.$$ 

Of course, Simpson's rule is of the form $$\int_a^b g(x) \: \text{d}x \approx \frac{b-a}{6}\left( g(a) +4g\!\left( \frac{a+b}{2} \right) +g(b) \right),$$ but in this case $f(x) \to \infty$ as $x \to 1$. I'm not sure if there is any kind of work around here, any help would be great!
 A: By integrating by parts, using
$$
(e^x)'=e^x, \qquad \left(\arcsin x \right)'=\frac{1}{\sqrt{1-x^2}},\qquad -1<x<1,
$$ one gets
$$
\int_0^1 \frac{e^x}{\sqrt{1-x^2}} \: dx=\left.e^x \frac{}{}\arcsin x\right|_0^1-\int_0^1e^x \frac{}{}\arcsin x \:dx
$$ and the latter integral is more suitable for the Simpson rule, the function $\arcsin(\cdot)$ being bounded over $[0,1]$.
A: Since by Euler's Beta function
$$ \int_{0}^{1}\frac{x^k\,dx}{\sqrt{1-x^2}} = \frac{1}{2}\int_{0}^{1}x^{(k-1)/2}(1-x)^{-1/2}\,dx = \frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\!\left(\tfrac{k+1}{2}\right)}{\Gamma\!\left(\tfrac{k+2}{2}\right)}\tag{1}$$
we simply have:
$$ \int_{0}^{1}\frac{e^x}{\sqrt{1-x^2}}\,dx = \sum_{k\geq 0}\frac{\sqrt{\pi}}{2k!}\cdot\frac{\Gamma\!\left(\tfrac{k+1}{2}\right)}{\Gamma\!\left(\tfrac{k+2}{2}\right)}\tag{2}$$
and the last hypergeometric series (that equals $\frac{\pi}{2}\left(I_0(1)+L_0(1)\right)$ in terms of Bessel and Struve functions) is fast-convergent. It is not difficult, either, to find a generalized continued fraction converging to the RHS of $(2)$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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As $\ds{x \to 1^{-}}$, $\ds{\root{1 - x^{2}} \sim 2^{1/2}\root{1 - x}}$ which suggests the variable change
  $$
\root{1 - x} \equiv t\quad \implies\quad x = 1 - t^{2}
$$
  This serves to the purpose of 'avoid' the integrable singularity as
  $\ds{x \to 1^{-}}$:

\begin{align}
\int_{0}^{1}{\expo{x} \over \root{1 - x^{2}}}\,\dd x &
\,\,\,\stackrel{x\ =\ 1 - t^{2}}{=}\,\,\,
\int_{1}^{0}\pars{-2\,{\expo{1 - t^{2}} \over \root{2-t^{2}}}}\dd t =
2\expo{}\int_{0}^{1}{\expo{-t^{2}} \over \root{2 - t^{2}}}
\end{align}
In using your announced Simpson's Rule:
\begin{align}
\int_{0}^{1}{\expo{x} \over \root{1 - x^{2}}}\,\dd x & =
2\expo{}\bracks{{1 - 0 \over 6}\pars{{\expo{-0^{2}} \over \root{2 - 0^{2}}} +
4\,{\expo{-1/4} \over \root{2 - 1/4}} + {\expo{-1^{2}} \over \root{2 - 1^{2}}}}}
\\[5mm] & =
{\root{2}\expo{} \over 6} +  {8\root{7}\expo{3/4} \over 21} + {1 \over 3}
\approx 3.10\color{#f00}{77\ldots}
\end{align}

The exact result $\ds{\pars{~\approx 3.10437901\ldots~}}$involves Bessel functions. The above Simpson's Rule yields a relative error
  $\ds{\approx 0.1094}$ %.

