Answer
$=\frac{1}{2}\ln5$
Work Step by Step
We are given $\int^{2}_{0}\int^{1}_{1/2}\frac{1}{y^{2}+1}dydx=\int^{2}_{0}(\int^{1}_{1/2}\frac{1}{y^{2}+1}dy)dx$
solve the inner integral
$\int^{1}_{1/2}\frac{1}{y^{2}+1}dy$
$=\frac{1}{y^{2}+1}\int^{1}_{1/2}dy$
$=\frac{1}{y^{2}+1}[x]^{1}_{1/2}$
$=\frac{1}{2y^{2}+2}$
then
$\int^{2}_{0}\int^{1}_{1/2}\frac{1}{y^{2}+1}dydx=\int^{2}_{0}\frac{1}{2y^{2}+2}dx$
intergrating
$[\frac{1}{2}\ln|2y^{2}+2|]^{2}_{0}$
$=\frac{1}{2}(\ln10-\ln2)$
$=\frac{1}{2}\ln5$
