Answer
$$\frac{{{y^2}}}{2} - \ln \left| y \right| + \frac{1}{2}\ln \left( {{y^2} + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{y^4} + {y^2} - 1}}{{{y^3} + y}}} dy \cr
& \cr
& {\text{Apply long division to the integrand }}\frac{{{y^4} + {y^2} - 1}}{{{y^3} + y}} \cr
& \frac{{{y^4} + {y^2} - 1}}{{{y^3} + y}} = y - \frac{1}{{{y^3} + y}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{{y^4} + {y^2} - 1}}{{{y^3} + y}}} dy = \int {\left( {y - \frac{1}{{{y^3} + y}}} \right)} dy \cr
& = \int y dy - \int {\frac{1}{{{y^3} + y}}} dy \cr
& = \frac{{{y^2}}}{2} - \int {\frac{1}{{{y^3} + y}}} dy \cr
& \cr
& {\text{Solve the integral by the method of partial fractions.}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{1}{{{y^3} + y}} = \frac{1}{{y\left( {{y^2} + 1} \right)}} = \frac{A}{y} + \frac{{By + C}}{{{y^2} + 1}} \cr
& {\text{Multiplying by }}y\left( {{y^2} + 1} \right){\text{, we have}} \cr
& 1 = A\left( {{y^2} + 1} \right) + \left( {By + C} \right)y \cr
& 1 = A{y^2} + A + B{y^2} + Cy \cr
& 1 = \left( {A{y^2} + B{y^2}} \right) + Cy + A \cr
& {\text{equating coefficients}} \cr
& A + B = 0,\,\,\,C = 0{\text{ and }}A = 1,\,\,\,B = - 1 \cr
& \frac{1}{{y\left( {{y^2} + 1} \right)}} = \frac{A}{y} + \frac{{By + C}}{{{y^2} + 1}} = \frac{1}{y} - \frac{y}{{{y^2} + 1}} \cr
& \cr
& \int {\frac{1}{{{y^3} + y}}} dy = \int {\left( {\frac{1}{y} - \frac{y}{{{y^2} + 1}}} \right)} dy \cr
& \int {\frac{1}{{{y^3} + y}}} dy = \ln \left| y \right| - \frac{1}{2}\ln \left( {{y^2} + 1} \right) + C \cr
& {\text{Then}}{\text{,}} \cr
& \frac{{{y^2}}}{2} - \int {\frac{1}{{{y^3} + y}}} dy = \frac{{{y^2}}}{2} - \ln \left| y \right| + \frac{1}{2}\ln \left( {{y^2} + 1} \right) + C \cr} $$
