QuestionEvaluate the integral $$\int_{1}^{\sqrt{9}} \frac{1}{1+x^{2}} d x$$

Studdy Solution
Compute the definite integral:
Subtract the evaluated lower limit from the evaluated upper limit:
$\int_{1}^{3} \frac{1}{1+x^{2}} \, dx = \arctan(3) - \arctan(1)$
The value of the integral is:
$\boxed{\arctan(3) - \frac{\pi}{4}}$