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

Question

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

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Answer 1 · 2018-06-22T00:23:32

1.2315

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