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