## Thomas' Calculus 13th Edition

$$\frac{{{t^2}}}{2} - 2t + \ln \left( {{t^2} + 1} \right) - 2{\tan ^{ - 1}}\left( t \right) + C$$
\eqalign{ & \int {\frac{{{t^3} - 2{t^2} + 3t - 4}}{{{t^2} + 1}}} dt \cr & {\text{simplify the improper fraction }}\frac{{{t^3} - 2{t^2} + 3t - 4}}{{{t^2} + 1}} \cr & \frac{{{t^3} - 2{t^2} + 3t - 4}}{{{t^2} + 1}} = t - 2 + \frac{{2t - 2}}{{{t^2} + 1}} \cr & = t - 2 + \frac{{2t}}{{{t^2} + 1}} - \frac{2}{{{t^2} + 1}} \cr & \int {\frac{{{t^3} - 2{t^2} + 3t - 4}}{{{t^2} + 1}}} dt = \int {\left( {t - 2 + \frac{{2t}}{{{t^2} + 1}} - \frac{2}{{{t^2} + 1}}} \right)} dt \cr & {\text{sum rule for integrals}} \cr & = \int t dt - \int 2 dt + \int {\frac{{2t}}{{{t^2} + 1}}} dt - \int {\frac{2}{{{t^2} + 1}}} dt \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 & = \frac{{{t^2}}}{2} - 2t + \ln \left( {{t^2} + 1} \right) - 2{\tan ^{ - 1}}\left( t \right) + C \cr}