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#

1 Answer
Jun 22, 2018

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