Answer
$\frac{\pi}{4}-\frac{1}{2}ln2$
Work Step by Step
$\int\int_{D} \frac{1}{1+x^{2}}dA=\int_{0}^{1}\int_{x} ^{1} \frac{1}{1+x^{2}}dydx$
$=\int_{0}^{1} \frac{y}{1+x^{2}}]_{x} ^{1}dx$
$=\int_{0} ^{1}\frac{1-x}{1+x^{2}}dx$
$=\int_{0} ^{1}\frac{1}{1+x^{2}}dx-\int_{0} ^{1}\frac{x}{1+x^{2}}dx$
$=[tan^{-1}x]_{0} ^{1}-\frac{1}{2}\int_{0} ^{1}\frac{2x}{1+x^{2}}dx$
$=[tan^{-1}(1)-tan^{-1}(0)]-\frac{1}{2}[ln{(1+x^{2})}]_{0} ^{1}$
$=\frac{\pi}{4}-\frac{1}{2}ln2$
