Nice way to solve $\int\int \frac{1}{1-(xy)^2} dydx$? This is something I've been thinking about lately;
$$\int_0^1 \int_0^1 \frac{1}{1-(xy)^2} dydx$$ 
Solutions I've read involve making the substitutions: $x= \frac{sin(u)}{cos(v)}$ and $y= \frac{sin(v)}{cos(u)}$. This reduces the integral to the area of a right triangle with both legs of length $\frac{\pi}{2}$. My problem is that coming up with this substitution is not at all obvious to me, and realizing how the substitution distorts the unit square into a right triangle seems to require a lot of reflection. My approach without fancy tricks involves letting $u = xy$ and then the integral "simplifies" accordingly:
$\begin{align*} \int_0^1 \int_0^1 \frac{1}{1-(xy)^2} dydx &= \int_0^1\frac{1}{x}\int_0^x \frac{1}{1-u^2}dudx\\
&= \int_0^1\frac{1}{2x}\int_0^x \frac{1}{1-u}+\frac{1}{1+u}dudx\\
&= \int_0^1\frac{1}{2x}ln\left(\frac{1+x}{1-x}\right)dx
 \end{align*}$ 
If I've done everything right this should be $\frac{\pi^2}{8}$ but I haven't figured out how to solve it.
 A: Here is an alternative derivation. The form of the integrand suggest expanding it a geometrical series
$$\frac{1}{1-(xy)^2} = \sum_{n=0}^\infty x^{2n}y^{2n}$$
Now integrating term by term we get
$$\int_0^1\int_0^1\frac{{\rm d}x{\rm d}y}{1-(xy)^2} = \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$
and from the well known result $\frac{\pi^2}{6} = \sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=0}^\infty\frac{1}{(2n+1)^2} + \sum_{n=1}^\infty\frac{1}{(2n)^2}$ we get
$$\int_0^1\int_0^1\frac{{\rm d}x{\rm d}y}{1-(xy)^2} = \frac{\pi^2}{8}$$
A: If you haven't an idea, just start:
$$\int_{0}^{1}\int_{0}^{1}\frac{1}{1-(xy)^2}\space\text{d}y\text{d}x=$$
$$\int_{0}^{1}\left[\int_{0}^{1}\frac{1}{1-(xy)^2}\space\text{d}y\right]\text{d}x=$$ 

For the integrand $u=xy$ and $\text{d}u=x\space\text{d}y$.
This gives a new lower bound $u=x\cdot0=0$ and upper bound $u=x\cdot1=x$:

$$\int_{0}^{1}\left[\frac{1}{x}\int_{0}^{x}\frac{1}{1-u^2}\space\text{d}u\right]\text{d}x=$$ 
$$\int_{0}^{1}\left[\frac{1}{x}\left[\tanh^{-1}\left(u\right)\right]_{0}^{x}\right]\text{d}x=$$ 
$$\int_{0}^{1}\left[\frac{1}{x}\left(\tanh^{-1}\left(x\right)-\tanh^{-1}\left(0\right)\right)\right]\text{d}x=$$ 
$$\int_{0}^{1}\left[\frac{1}{x}\left(\tanh^{-1}\left(x\right)-0\right)\right]\text{d}x=$$ 
$$\int_{0}^{1}\left[\frac{\tanh^{-1}\left(x\right)}{x}\right]\text{d}x=$$ 
$$\frac{1}{2}\left[\text{Li}_2(x)-\text{Li}_2(-x)\right]_{0}^{1}=$$ 
$$\frac{1}{2}\left(\left(\text{Li}_2(1)-\text{Li}_2(-1)\right)-\left(\text{Li}_2(0)-\text{Li}_2(-0)\right)\right)=$$ 
$$\frac{1}{2}\left(\frac{\pi^2}{4}-0\right)=$$ 
$$\frac{1}{2}\left(\frac{\pi^2}{4}\right)=\frac{\pi^2}{8}$$ 
A: You can also do this, and this one does not involve any fancy tricks: 
Consider the double integral: $$I=\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{(1+x)(x+y^2)}dydx.$$ 
Integrate with respect to $y$ first and recognize this is :
$$I=\int_{0}^{\infty} \frac{1}{1+x}\lim_{y \rightarrow \infty} \frac{\arctan{\frac{y}{\sqrt{x}}}}{\sqrt {x}} dx=\int_{0}^{\infty}\frac{\frac{\pi}{2}}{\sqrt{x}{(1+x)}}dx.$$ Now if you apply the transformation $u=\sqrt{x}, du=\frac{1}{2\sqrt{x}}dx,$ you get $$I=\int_{0}^{\infty} \frac{\pi}{(1+u^2)}du=\frac{\pi^2}{2}.$$
On the other hand, reverse the order of integration in $I$ as such: $$I=\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{(1+x)(x+y^2)}dxdy.$$
Now integrate with respect to $x$ using partial fractions as such:
$$\frac{1}{(1+x)(x+y^2)}=\frac{1}{y^2-1} \left(\frac{1}{1+x}-\frac{1}{x+y^2}\right).$$
$$I=\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{y^2-1} \left(\frac{1}{1+x}-\frac{1}{x+y^2}\right)dxdy=\int_{0}^{\infty}\lim_{x\rightarrow \infty}\frac{\ln(1+x)-\ln(x+y^2)}{y^2-1}dy=\int_{0}^{\infty}\frac{\ln(y^2)}{y^2-1}dy=\int_{0}^{\infty}\frac{2\ln(y)}{y^2-1}dy.$$
Now consider $$J=\int_{0}^{1}\frac{\ln(y)}{y^2-1}dy.$$ A u-substitution $y=\frac{1}{z}, dy=\frac{-1}{z^2} dz$ tells us: 
$$J=\int_{1}^{\infty}\frac{\ln(z)}{z^2-1}dz.$$  This tells us: $$J=\frac{I}{4}=\frac{\pi^2}{8}.$$ Now let us make the substitution $z=\frac{1-t}{1+t}, dz=\frac{-2}{(t+1)^2}dt.$ We have:
$$J=\frac{\pi^2}{8}=\int_{0}^{1} \frac{\ln(1+t)-\ln(1-t)}{2t} dt,$$ the integral you arrived at.  
