Short way to evaluate $\int_{0}^{1}\ln(\frac{a-x^2}{a+x^2})\cdot\frac{dx}{x^2\sqrt{1-x^2}}$ I would like to evaluate this integral, $$I=\int_{0}^{1}\ln\left(\frac{a-x^2}{a+x^2}\right)\cdot\frac{\mathrm dx}{x^2\sqrt{1-x^2}}$$
An approach:
1. Using integration by parts
$u=\ln\left(\frac{a-x^2}{a+x^2}\right)$, $du=\frac{x^2+a}{x^2-a}\left[\frac{2x}{a+x^2}+\frac{2x(a-x^2)}{a+x^2}\right]dx$
$dv=\frac{dx}{x^2\sqrt{1-x^2}}$, $v=-\frac{\sqrt{1-x^2}}{x}$
$$I=2\int_{0}^{1}\frac{\sqrt{1-x^2}}{x^2+a}dx-2\int_{0}^{1}\frac{\sqrt{1-x^2}}{x^2-a}dx=2I_1-2I_2$$
Integral $I_1$: Make a substitution of: $x=\sin(v)$, $v=\arcsin(x)$, $dx=\cos(v)dv$
$$I_1=\int_{0}^{\pi/2}\frac{\cos^2(v)}{a+\sin^2(v)}dv$$
Using trig to rewrite: $$I_1=\int_{0}^{\pi/2}\sec^2(v)\frac{dv}{[1+\tan^2(v)][(a+1)\tan^2(v)+a]}$$
Make another substitiution of: $v=\tan(y)$, $dy=\frac{1}{\sec^2(v)}dv$ and using partial fraction decomp:
$$I_1=(a+1)\int_{0}^{\infty}\frac{dy}{a+(a+1)y^2}-\frac{\pi}{2}=(a+1)I_3-\frac{\pi}{2}$$
With this substitution: $t=\frac{\sqrt{a+1}}{\sqrt{a}}y$, $dy=\frac{\sqrt{a}}{\sqrt{a+1}}dt$
$$I_3=\frac{1}{\sqrt{a(a+1)}}\cdot \frac{\pi}{2}$$
$$I_1=\frac{\pi}{2}\left(\frac{\sqrt{a+1}}{\sqrt{a}}-1\right)$$
In the same manner, $$I_2=\frac{\pi}{2}\left(\frac{\sqrt{a-1}}{\sqrt{a}}-1\right)$$ 
Finally, $$I=\pi\left(\frac{\sqrt{a-1}-\sqrt{a+1}}{\sqrt{a}}\right)$$
I am looking for another short approach of evaluating integral $I$
 A: Assuming $a>1$, the original integral equals
$$ \frac{1}{2}\int_{0}^{1}\log\left(\frac{a-x}{a+x}\right)\frac{dx}{x\sqrt{x(1-x)}}\,dx=-\int_{0}^{1}\frac{\text{arctanh}(x/a)}{x\sqrt{x(1-x)}}\,dx \tag{1}$$
or
$$ -\sum_{n\geq 0}\int_{0}^{1}\frac{x^{2n-1/2}}{(2n+1)a^{2n+1}\sqrt{1-x}}\,dx=-\sum_{n\geq 0}\frac{\pi\binom{4n}{2n}}{(2n+1)16^n a^{2n+1}}\tag{2}$$
hence it is enough to recall the generating function for Catalan numbers to recover the wanted identity.

It is worth noticing that the Maclaurin series of the squared arcsine allows to compute the similar integral
$$ \int_{0}^{1}\log\left(\frac{a^2+x^2}{a^2-x^2}\right)\frac{dx}{\color{red}{x}\sqrt{1-x^2}} $$
or the hypergeometric function $\phantom{}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},1,1;\tfrac{3}{4},\tfrac{5}{4},\tfrac{3}{2};z\right)$ in terms of $\arcsin^2$ and $\text{arcsinh}^2$, in a very similar and efficient way.
A: Under $x\to\sin x$, one has
$$I=\int_{0}^{1}\ln\left(\frac{a-x^2}{a+x^2}\right)\cdot\frac{\mathrm dx}{x^2\sqrt{1-x^2}}=\int_{0}^{\pi/2}\ln\left(\frac{a-\sin^2x}{a+\sin^2x}\right)\cdot\frac{\mathrm dx}{\sin^2x}.$$
Let
$$I(k)=\int_{0}^{\pi/2}\ln\left(\frac{a-k\sin^2x}{a+k\sin^2x}\right)\cdot\frac{\mathrm dx}{\sin^2x}$$
and then
\begin{eqnarray}
I'(k)&=&\int_{0}^{\pi/2}\left(-\frac1{a-k\sin^2x}-\frac1{a+k\sin^2x}\right)\mathrm dx\\
&=&-\frac{\pi}{2}\left(\frac1{\sqrt{a(a+k)}}+\frac1{\sqrt{a(a-k)}}\right).
\end{eqnarray}
So
$$ I=I(1)=-\frac{\pi}{2\sqrt a}\int_0^1\left(\frac1{\sqrt{a+k}}-\frac1{\sqrt{a-k}}\right)\mathrm dk=-\frac{\pi}{\sqrt a}(\sqrt{a+1}-\sqrt{a-1}).$$
