Question

Help me please ! ASAP . . . .

Answer 1

`\textbf{Answer:}`

`\medskip\\ewline{\int\limits_(0)^(\infty){\bigl((1)/(y^2+1)\bigr)tan^(-1)(x)\mid_(x=0)^(x=\infty)}dy=(\pi^2)/(4)}`

`\medskip\\ewline\textbf{Step-by-step explanation:}`

`\medskip\\ewline\text{It's been a very long time since I've seen a calculus question.}\\ewline{\text{It might be because I haven't visited this site in a while.}}`
`\text{Anyway, on to the answer.}`

`\medskip\\ewline\text{When doing 2 integrals, work inside to out and treat the variables}\\ewline\text{ that aren't the ones being integrated as constants and proceed like normal.}`
`\medskip\medskip\\ewline\int\limits_(0)^(\infty){\int\limits_(0)^(\infty){(1)/((x^2+1)(y^2+1))}dx}dy`

`\\ewline{\text{Work inner to outer. Therefore, treat all y's as constants for now.}}`
`\\ewline{\text{By the integration formula: }\int(1)/(x^2+a^2)dx=(1)/(a)tan^(-1)((x)/(a))+C}`

`\\ewline{\int\limits_(0)^(\infty){\int\limits_(0)^(\infty){(1)/((x^2+1)(y^2+1))}dx}dy=}`

`\\ewline{\int\limits_(0)^(\infty){\bigl((1)/(y^2+1)\bigr)tan^(-1)(x)\mid\limits_(x=0)^(x=\infty)}dy=}`

`\\ewline{\int\limits_(0)^(\infty){\bigl((1)/(y^2+1)\bigr)(tan^(-1)(\infty)-tan^(-1)(0))}dy=}`

`\\ewline{\int\limits_(0)^(\infty){\bigl((1)/(y^2+1)\bigr)((\pi)/(2)-0)}dy=}`

`\\ewline{\int\limits_(0)^(\infty){\bigl((1)/(y^2+1)\bigr)((\pi)/(2))}dy}`

`\medskip\\ewline\text{Using the same integration formula as before.}`

`\medskip\\ewline{\int\limits_(0)^(\infty){\bigl((1)/(y^2+1)\bigr)((\pi)/(2))}dy=}`

`\\ewline{((\pi)/(2))tan^(-1)(y)\mid\limits_(y=0)^(y=\infty)=}`

`\\ewline{((\pi)/(2))(tan^(-1)(\infty)-tan^(-1)(0))=}`

`\\ewline{((\pi)/(2))((\pi)/(2)-0)=}`

`\\ewline{((\pi)/(2))((\pi)/(2))=}`

`\\ewline{(\pi^2)/(4)}`

`\bigskip\\ewline{\text{Therefore, the answer is }(\pi^2)/(4)}`