## Calculus: Early Transcendentals (2nd Edition)

$= {\tan ^{ - 1}}y - \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\,\left( {\frac{y}{{\sqrt 2 }}} \right) + C$
$\begin{gathered} \int_{}^{} {\frac{{dy}}{{\,\left( {{y^2} + 1} \right)\,\left( {{y^2} + 2} \right)}}} \hfill \\ \hfill \\ Using\,\,partial\,fraction \hfill \\ \hfill \\ \frac{1}{{\,\left( {{y^2} + 1} \right)\,\left( {{y^2} + 2} \right)}} = \frac{{Ay + B}}{{{y^2} + 1}} + \frac{{Cy + D}}{{{y^2} + 2}} \hfill \\ \hfill \\ {\text{find}}\,\,{\text{the}}\,\,{\text{constants}} \hfill \\ \hfill \\ A = 1,{\text{ B}} = 0,{\text{ C}} = - 1,{\text{ }}D = 0 \hfill \\ \hfill \\ {\text{Therefore}}{\text{,}} \hfill \\ \hfill \\ \frac{1}{{\,\left( {{y^2} + 1} \right)\,\left( {{y^2} + 2} \right)}} = \frac{1}{{{y^2} + 1}} - \frac{1}{{{y^2} + 2}} \hfill \\ \hfill \\ \int_{}^{} {\frac{1}{{\,\left( {{y^2} + 1} \right)\,\left( {{y^2} + 2} \right)}}dy} = \int_{}^{} {\,\left( {\frac{1}{{{y^2} + 1}} - \frac{1}{{{y^2} + 2}}} \right)dy} \hfill \\ \hfill \\ {\text{integrate}} \hfill \\ \hfill \\ = {\tan ^{ - 1}}y - \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\,\left( {\frac{y}{{\sqrt 2 }}} \right) + C \hfill \\ \end{gathered}$