Review on double integral Forgot how to do these :
$$\displaystyle\int_0^1\int_0^1\frac{\text{d}x\text{d}y}{1-x^2y^2}$$
 A: Consider the integral over $x$:
$$\int_0^1 dx \; \frac{1}{1-x^2 y^2} $$
Make the substitution $x = \sin{\theta}/y$, $dx = \cos{\theta}/y \: d \theta$:
$$\begin{align} &= \frac{1}{y} \int_0^{\arcsin{y}} d \theta \sec{\theta} \\ &=  \frac{1}{y} [\log{(\sec{\theta} + \tan{\theta})]_{0}^{\arcsin{y}}} \\ &= \frac{1}{2 y} \log{\left ( \frac{1+y}{1-y} \right )} \end{align} $$
Now you can do the integral over $y$:
$$\begin{align} \int_0^1 dy \: \int_0^1 dx \: \frac{1}{1-x^2 y^2} &= \frac{1}{2} \int_0^1 dy \: \frac{1}{y} \log{\left ( \frac{1+y}{1-y} \right )}\\ &= \sum_{n=0}^{\infty} \int_0^1 dy \: \frac{y^{2 n}}{2 n+1} \\ &= \sum_{n=0}^{\infty} \frac{1}{(2 n+1)^2}\\ &= \frac{\pi^2}{8} \\\\  \end{align} $$
A: *

*Note that your area $0\leq x\leq 1, 0\leq y\leq 1$ is a rectangular area so you can divide the integral respect to $x$ and $y$ as: $$\int_0^1~dx\times\int_0^1~f dy$$

*We know that $\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\ln|\frac{x+a}{x-a}|+C$.

*Write $\frac{1}{1-x^2y^2}$ as $\frac{(1/y^2)}{\left(\frac{1}{y^2}-x^2\right)}$ and then assume $1/y^2=a^2$ temporarily. 
So first solve $$A=\int_0^1\frac{a^2}{a^2-x^2}~dx$$ and then solve $$\int_0^1Ady$$ 
A: Here is another way.
$$\dfrac1{1-x^2y^2} = \sum_{k=0}^{\infty} x^{2k} y^{2k}$$
Hence,
\begin{align}
\int_0^1 \int_0^1 \dfrac{dx dy}{1-x^2y^2} & = \int_0^1 \int_0^1 \sum_{k=0}^{\infty} x^{2k} y^{2k} dxdy = \sum_{k=0}^{\infty} \int_0^1 \int_0^1 x^{2k} y^{2k} dxdy\\
& = \sum_{k=0}^{\infty} \left(\int_0^1 x^{2k} dx \right)^2 = \sum_{k=0}^{\infty} \dfrac1{(2k+1)^2} = \dfrac34 \zeta(2) = \dfrac{\pi^2}8
\end{align}
