Answer
\[\frac{{{y^2}}}{2} - \frac{1}{2}\ln \left( {{y^2} + 1} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dy}}{{{y^{ - 1}} + {y^{ - 3}}}}} \,\, = \int_{}^{} {\frac{{{y^3}dy}}{{{y^2} + 1}}} \hfill \\
\hfill \\
{\text{using}}\,\,\,{\text{the}}\,\,{\text{long}}\,\,\,{\text{division}} \hfill \\
\hfill \\
\int_{}^{} {\,\left( {y - \frac{y}{{{y^2} + 1}}} \right)} dy \hfill \\
\hfill \\
split\,\,the\,\,{\text{integrand}} \hfill \\
\hfill \\
\int_{}^{} {\,y} dy - \int_{}^{} {\,\frac{y}{{{y^2} + 1}}} dy \hfill \\
\hfill \\
or \hfill \\
\hfill \\
\int_{}^{} {\,y} dy - \frac{1}{2}\int_{}^{} {\,\frac{{2y}}{{{y^2} + 1}}} dy \hfill \\
\hfill \\
{\text{integrate}} \hfill \\
\hfill \\
\frac{{{y^2}}}{2} - \frac{1}{2}\ln \left( {{y^2} + 1} \right) + C \hfill \\
\end{gathered} \]