[tex]\textbf{Answer:}[/tex]
[tex]\medskip\newline{\int\limits_{0}^{\infty}{\bigl(\frac{1}{y^2+1}\bigr)tan^{-1}(x)\mid_{x=0}^{x=\infty}}dy=\frac{\pi^2}{4}}[/tex]
[tex]\medskip\newline\textbf{Step-by-step explanation:}[/tex]
[tex]\medskip\newline\text{It's been a very long time since I've seen a calculus question.}\newline{\text{It might be because I haven't visited this site in a while.}}[/tex][tex]\text{Anyway, on to the answer.}[/tex]
[tex]\medskip\newline\text{When doing 2 integrals, work inside to out and treat the variables}\newline\text{ that aren't the ones being integrated as constants and proceed like normal.}[/tex][tex]\medskip\medskip\newline\int\limits_{0}^{\infty}{\int\limits_{0}^{\infty}{\frac{1}{(x^2+1)(y^2+1)}}dx}dy[/tex]
[tex]\newline{\text{Work inner to outer. Therefore, treat all y's as constants for now.}}[/tex][tex]\newline{\text{By the integration formula: }\int\frac{1}{x^2+a^2}dx=\frac{1}{a}tan^{-1}(\frac{x}{a})+C}[/tex]
[tex]\newline{\int\limits_{0}^{\infty}{\int\limits_{0}^{\infty}{\frac{1}{(x^2+1)(y^2+1)}}dx}dy=}[/tex]
[tex]\newline{\int\limits_{0}^{\infty}{\bigl(\frac{1}{y^2+1}\bigr)tan^{-1}(x)\mid\limits_{x=0}^{x=\infty}}dy=}[/tex]
[tex]\newline{\int\limits_{0}^{\infty}{\bigl(\frac{1}{y^2+1}\bigr)(tan^{-1}(\infty)-tan^{-1}(0))}dy=}[/tex]
[tex]\newline{\int\limits_{0}^{\infty}{\bigl(\frac{1}{y^2+1}\bigr)(\frac{\pi}{2}-0)}dy=}[/tex]
[tex]\newline{\int\limits_{0}^{\infty}{\bigl(\frac{1}{y^2+1}\bigr)(\frac{\pi}{2})}dy}[/tex]
[tex]\medskip\newline\text{Using the same integration formula as before.}[/tex]
[tex]\medskip\newline{\int\limits_{0}^{\infty}{\bigl(\frac{1}{y^2+1}\bigr)(\frac{\pi}{2})}dy=}[/tex]
[tex]\newline{(\frac{\pi}{2})tan^{-1}(y)\mid\limits_{y=0}^{y=\infty}=}[/tex]
[tex]\newline{(\frac{\pi}{2})(tan^{-1}(\infty)-tan^{-1}(0))=}[/tex]
[tex]\newline{(\frac{\pi}{2})(\frac{\pi}{2}-0)=}[/tex]
[tex]\newline{(\frac{\pi}{2})(\frac{\pi}{2})=}[/tex]
[tex]\newline{\frac{\pi^2}{4}}[/tex]
[tex]\bigskip\newline{\text{Therefore, the answer is }\frac{\pi^2}{4}}[/tex]
