Answer
$$x + \ln \left( {{x^2} + 9} \right) - \frac{{10}}{3}{\tan ^{ - 1}}\left( {\frac{x}{3}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2} + 2x - 1}}{{{x^2} + 9}}} dx \cr
& {\text{simplify the improper fraction }}\frac{{{x^2} + 2x - 1}}{{{x^2} + 9}} \cr
& \frac{{{x^2} + 2x - 1}}{{{x^2} + 9}} = 1 + \frac{{2x - 10}}{{{x^2} + 9}} \cr
& \frac{{{x^2} + 2x - 1}}{{{x^2} + 9}} = 1 + \frac{{2x}}{{{x^2} + 9}} - \frac{{10}}{{{x^2} + 9}} \cr
& {\text{then}} \cr
& \int {\frac{{{x^2} + 2x - 1}}{{{x^2} + 9}}} dx = \int {\left( {1 + \frac{{2x}}{{{x^2} + 9}} - \frac{{10}}{{{x^2} + 9}}} \right)} dx \cr
& {\text{sum rule for integrals gives}} \cr
& = \int {dx} + \int {\frac{{2x}}{{{x^2} + 9}}} dx - \int {\frac{{10}}{{{x^2} + 9}}} dx \cr
& {\text{integrate by using the formulas}} \cr
& \int {\frac{{du}}{u}} = \ln \left| u \right| + C{\text{ and }}\int {\frac{1}{{{u^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + C \cr
& {\text{then}} \cr
& = x + \ln \left( {{x^2} + 9} \right) - \frac{{10}}{3}{\tan ^{ - 1}}\left( {\frac{x}{3}} \right) + C \cr} $$
