This is a question I came across and I cannot find the answer.
By using a substitution involving the tangent function, show that $$\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx=\frac{\pi}{4}$$
My attempt
I use trig substitution, by saying $$\tan(\theta)=\frac{y}{x}$$ which means $$x\sec^2(\theta)\,d\theta=dy$$ Also, it should be noted that because of this $$x\sec(\theta)=\sqrt{x^2+y^2}$$ $$x^4\sec^4(\theta)=(x^2+y^2)^2 $$ Thus, when I substitute this information into the integral, I get $$\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2-(x^2\tan^2(\theta))}{x^4\sec^4(\theta)}{x\sec^2(\theta)} \, d\theta \,dx$$ Then, this simplifies to $$\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2(1-\tan^2(\theta))}{x^4\sec^4(\theta)}{x\sec^2(\theta)}\,d\theta \,dx$$ $$\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2\sec^2(\theta)}{x^4\sec^4(\theta)}{x\sec^2(\theta)}\,d\theta \,dx=\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})}{\frac{1}{x}}\,d\theta \,dx$$ which leads to $$\int_0^1 \left[\frac \theta x \right]_0^{\arctan\left(\frac{1}{x}\right)} \,dx = \int_0^1 \frac{\arctan\left(\frac{1}{x}\right)}{x} \, dx_{(3)} $$ At this point I am stuck. How do I evaluate this integral. Am I on the right path? Wolfram Alpha gives an answer other than $\frac{\pi}{4}$ for (3), so I am not sure where I am wrong.