Answer
$$\frac{{{y^3}}}{3} - y + {\tan ^{ - 1}}y + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{y^4}}}{{{y^2} + 1}}dy} \cr
& {\text{expanding integrand}} \cr
& = \int {\left( {{y^2} - 1 + \frac{1}{{{y^2} + 1}}} \right)dy} \cr
& {\text{sum rule}} \cr
& = \int {{y^2}dy} - \int {dy} + \int {\frac{1}{{{y^2} + 1}}dy} \cr
& = \int {{y^2}dy} - \int {dy} + \int {\frac{1}{{{y^2} + 1}}dy} \cr
& {\text{integrating}} \cr
& {\text{use power rule}} \cr
& {\text{use }}\int {\frac{{du}}{{{u^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\frac{u}{a} + c \cr
& = \frac{{{y^3}}}{3} - y + {\tan ^{ - 1}}y + C \cr} $$