Question

How do you do 2/(1+x^2+y^2) integration?

`int_0^2int_0^(-x/2+1)2/(1+x^2+y^2)dydx`

Answer 1

1.2315

Explanation:

You evaluate the double integral

`int_0^2int_0^(-x/2+1)2/(1+x^2+y^2)dydx`

by iterated integration, i.e. by first integrating over `y` for a constant `x` , and then integrating over `x`

We first evaluate

`int_0^(-x/2+1)2/(1+x^2+y^2)dy`

remembering that `x` is to be treated as a constant here, we see that this is

`2/sqrt(1+x^2)(tan^-1 (y/sqrt(1+x^2)))_0^{-x/2+1}`
`= 2/sqrt(1+x^2) tan^-1((-x/2+1)/sqrt(1+x^2))`

To complete, we need to evaluate

`int_0^2 2/sqrt(1+x^2) tan^-1((-x/2+1)/sqrt(1+x^2))dx`

This, however, can not be evaluated analytically. It is possible to evaluate this numerically, though, and the value comes out to be

1.2315