Answer
$$A \approx 2.680621$$
Work Step by Step
$$\eqalign{
& y = \frac{{3x + 2}}{{{x^2} + 9}} \cr
& {\text{The area of the region is given by}} \cr
& A = \int_0^5 {\frac{{3x + 2}}{{{x^2} + 9}}dx} \cr
& {\text{Split the numerator}} \cr
& A = \int_0^5 {\left( {\frac{{3x}}{{{x^2} + 9}} + \frac{2}{{{x^2} + 9}}} \right)dx} \cr
& A = \int_0^5 {\frac{{3x}}{{{x^2} + 9}}dx} + \int_0^5 {\frac{2}{{{x^2} + 9}}dx} \cr
& {\text{Rewrite integrands}} \cr
& A = \frac{3}{2}\int_0^5 {\frac{{2x}}{{{x^2} + 9}}dx} + 2\int_0^5 {\frac{1}{{{x^2} + 9}}dx} \cr
& {\text{Integrate using basic integration formulas}} \cr
& A = \frac{3}{2}\left[ {\ln \left( {{x^2} + 9} \right)} \right]_0^5 + 2\left[ {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{x}{3}} \right)} \right]_0^5 \cr
& A = \left[ {\frac{3}{2}\ln \left( {{x^2} + 9} \right) + \frac{2}{3}{{\tan }^{ - 1}}\left( {\frac{x}{3}} \right)} \right]_0^5 \cr
& A = \left[ {\frac{3}{2}\ln \left( {{5^2} + 9} \right) + \frac{2}{3}{{\tan }^{ - 1}}\left( {\frac{5}{3}} \right)} \right] - \left[ {\frac{3}{2}\ln \left( {{0^2} + 9} \right) + \frac{2}{3}{{\tan }^{ - 1}}\left( {\frac{0}{3}} \right)} \right] \cr
& A = \left[ {\frac{3}{2}\ln \left( {34} \right) + \frac{2}{3}{{\tan }^{ - 1}}\left( {\frac{5}{3}} \right)} \right] - \left[ {\frac{3}{2}\ln \left( 9 \right) + 0} \right] \cr
& A = \frac{3}{2}\ln \left( {34} \right) + \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{5}{3}} \right) - \frac{3}{2}\ln \left( 9 \right) \cr
& A \approx 2.680621 \cr} $$
