Series approximation to $\int_0^1\sqrt{\frac{2x+3}{2u^3-(2x+3)u^2+2x+1}}du$ I have figured out by graphing that, for small $x$:
$$
\int_0^1\sqrt{\frac{2x+3}{2u^3-(2x+3)u^2+2x+1}}du\approx\log(1/x)+\pi/2+O(x)
$$
However, I am unable to prove that this is the case.
As $x\to 0$ the integral diverges at $u=1$.  We can check this by factoring the limit of the integrand:
$$
\sqrt{\frac{3}{1+2u}}\frac{1}{1-u}
$$
This looks promising, since the integral of $1/u$ is $\log u$.  However, normally the integral converges: taking a series around $u=1$ we get:
$$
\sqrt{\frac{2x+3}{4x}\frac{1}{1-u}}+O(\sqrt{1-x})
$$
This means that when I take the series approximation of $x$ inside the integral, the resulting integral goes like $x^{-1/2}$, which is incorrect.  What step am I missing?
 A: I think the resulting behavior is indeed a log as $x \to 0$.  To see why, note that the denominator may be written as
$$f(u,x) = 2 u^3 - 3 u^2 + 1 + 2 x (1 - u^2)$$
This implies that $1-u$ is a factor of $f(u,x)$ independent of $x$.  In fact,
$$f(u,x) = (1-u)(1+2 x + (1+2 x) u - 2 u^2) $$
The zeroes of the quadratic are at
$$u_{\pm} = \frac{1}{4} \left(1+2 x \pm \sqrt{9+20 x+4 x^2}\right)$$
The behavior of these zeroes as $x \to 0$ is
$$u_+ = 1+\frac{4 x}{3}+O\left(x^2\right) $$
$$u_- = -\frac{1}{2}-\frac{x}{3}+O\left(x^2\right) $$
Now let's turn our attention to the integral.  Let's ignore the numerator, as it is merely a constant.  What we have is, as $x \to 0$,
$$\frac1{\sqrt{2}} \int_0^1 \frac{du}{\sqrt{(1-u) \left (\frac{1}{2}+\frac{x}{3}+u \right ) \left (1+\frac{4 x}{3}-u \right )}} $$
Now sub $u=1-v^2$ in the integral.  We get, as $x \to 0$ (we can neglect the $x/3$):
$$\sqrt{2} \int_0^1 \frac{dv}{\sqrt{ \left (\frac{3}{2}-v^2 \right )\left (\frac{4 x}{3}+v^2 \right )}} $$
The integral above is in fact an incomplete elliptic integral of the first kind (as defined by Wolfram Mathematica), equal to
$$\sqrt{\frac{3}{x}} \operatorname{F}{\left (\sin ^{-1}\left(\sqrt{\frac{2}{3}}\right),-\frac{9}{8 x} \right )} $$
The leading asymptotic behavior of this result as $x \to 0$, which I did not work out by hand, is
$$\sqrt{\frac{1}{3}} \left (\log{\frac1{x}}-\log {\left[\frac{1}{18} \left(2+\sqrt{3}\right)\right ]}\right ) $$
This seems to have the correct behavior when compared with the numerical integral for small $x$.
ADDENDUM
Interestingly enough, compare the result you proposed:
$$\log{\frac1{x}} + \frac{\pi}{2} \approx \log{\frac1{x}} +1.5708$$
with the analogous result resulting from my illustration:
$$\log{\frac1{x}}-\log {\left[\frac{1}{18} \left(2+\sqrt{3}\right)\right ]} \approx \log{\frac1{x}} +1.5734$$
so your hypothesis was not wildly off.
