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

## ${\int}_{0}^{2} {\int}_{0}^{- \frac{x}{2} + 1} \frac{2}{1 + {x}^{2} + {y}^{2}} \mathrm{dy} \mathrm{dx}$

Jun 22, 2018

1.2315

#### Explanation:

You evaluate the double integral

${\int}_{0}^{2} {\int}_{0}^{- \frac{x}{2} + 1} \frac{2}{1 + {x}^{2} + {y}^{2}} \mathrm{dy} \mathrm{dx}$

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

We first evaluate

${\int}_{0}^{- \frac{x}{2} + 1} \frac{2}{1 + {x}^{2} + {y}^{2}} \mathrm{dy}$

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

$\frac{2}{\sqrt{1 + {x}^{2}}} {\left({\tan}^{-} 1 \left(\frac{y}{\sqrt{1 + {x}^{2}}}\right)\right)}_{0}^{- \frac{x}{2} + 1}$
$= \frac{2}{\sqrt{1 + {x}^{2}}} {\tan}^{-} 1 \left(\frac{- \frac{x}{2} + 1}{\sqrt{1 + {x}^{2}}}\right)$

To complete, we need to evaluate

${\int}_{0}^{2} \frac{2}{\sqrt{1 + {x}^{2}}} {\tan}^{-} 1 \left(\frac{- \frac{x}{2} + 1}{\sqrt{1 + {x}^{2}}}\right) \mathrm{dx}$

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

1.2315