Answer
$0.108$
Work Step by Step
$\int_{3}^{6}\frac{1}{25+(x-3)^{2}}dx$
Notice that the denominator $25+(x-3)^{2}$ is similar to $arctan$. To solve this integral, use the formula to solve to $arctan$:
$\int\frac{du}{a^{2}+u^{2}}=\frac{1}{a}arctan(\frac{u}{a})+C$
$u=x-3$
$du=dx$
$a=5$
$[\frac{1}{5}arctan(\frac{x-3}{5})]_{3}^{6}$
$\frac{1}{5}[arctan(\frac{6-3}{5})-arctan\frac{3-3}{5}$
$\frac{1}{5}[arctan\frac{3}{5}-arctan0]$
$\approx0.108$
