Answer
$$\frac{{{x^3}}}{3} + \frac{1}{2}\ln \left( {{x^2} + 6x + 10} \right) - 3{\tan ^{ - 1}}\left( {x + 3} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^4} + 6{x^3} + 10{x^2} + x}}{{{x^2} + 6x + 10}}dx} \cr
& {\text{Rewrite the integrand}} \cr
& = \int {\left( {\frac{{{x^4} + 6{x^3} + 10{x^2}}}{{{x^2} + 6x + 10}} + \frac{x}{{{x^2} + 6x + 10}}} \right)dx} \cr
& = \int {\left( {\frac{{{x^2}\left( {{x^2} + 6x + 10} \right)}}{{{x^2} + 6x + 10}} + \frac{x}{{{x^2} + 6x + 10}}} \right)dx} \cr
& = \int {\left( {{x^2} + \frac{x}{{{x^2} + 6x + 10}}} \right)dx} \cr
& = \int {{x^2}} dx + \int {\frac{x}{{{x^2} + 6x + 10}}} dx \cr
& {\text{Completing the square for }}{x^2} + 6x + 10 \cr
& = \int {{x^2}} dx + \int {\frac{x}{{{x^2} + 6x + 9 + 1}}} dx \cr
& = \int {{x^2}} dx + \int {\frac{x}{{{{\left( {x + 3} \right)}^2} + 1}}} dx \cr
& {\text{Integrating }}\int {\frac{x}{{{{\left( {x + 3} \right)}^2} + 1}}} dx \cr
& {\text{let }}u = x + 3,{\text{ }}du = dx \cr
& \int {\frac{x}{{{{\left( {x + 3} \right)}^2} + 1}}} dx = \int {\frac{{u - 3}}{{{u^2} + 1}}} du \cr
& = \int {\frac{u}{{{u^2} + 1}}} du - \int {\frac{3}{{{u^2} + 1}}} \cr
& = \frac{1}{2}\ln \left( {{u^2} + 1} \right) - 3{\tan ^{ - 1}}u + C \cr
& = \frac{1}{2}\ln \left[ {{{\left( {x + 3} \right)}^2} + 1} \right] - 3{\tan ^{ - 1}}\left( {x + 3} \right) + C \cr
& = \frac{1}{2}\ln \left( {{x^2} + 6x + 10} \right) - 3{\tan ^{ - 1}}\left( {x + 3} \right) + C \cr
& ,{\text{ then}} \cr
& = \frac{{{x^3}}}{3} + \frac{1}{2}\ln \left( {{x^2} + 6x + 10} \right) - 3{\tan ^{ - 1}}\left( {x + 3} \right) + C \cr} $$