Evaluate, $I = \int {\frac{{{x^3} + 2x + 1}}{{{x^2} + 1}}} dx$
We have, $I = \int {\frac{{{x^3} + x + x + 1}}{{{x^2} + 1}}} dx$ = $ \int {\frac{{x({x^2} + 1) + x + 1}}{{{x^2} + 1}}} dx$
$ = \int {x + \frac{{x + 1}}{{{x^2} + 1}}dx} $
$ = \int {xdx + \int {\frac{x}{{{x^2} + 1}}} } dx + \int {\frac{{dx}}{{1 + {x^2}}}} $
$ = \frac{{{x^2}}}{2} + \frac{1}{2}\int {\frac{{2x}}{{{x^2} + 1}}} dx + {\tan ^{ - 1}}x + C$
$ = \frac{{{x^2}}}{2} + \frac{1}{2}\ln ({x^2} + 1) + {\tan ^{ - 1}}x + C$