Answer
$\frac{{{y^2}}}{4}\sqrt {4 + {y^4}} - \ln \left( {{y^2} + \sqrt {4 + {y^4}} } \right) + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{y^5}}}{{\sqrt {4 + {y^4}} }}} dy; \cr
& \int {\frac{{{y^5}}}{{\sqrt {4 + {y^4}} }}} dy = \int {\frac{{{{\left( {{y^2}} \right)}^2}y}}{{\sqrt {4 + {{\left( {{y^2}} \right)}^2}} }}} dy \cr
& {\text{Let }}u = {y^2} \Rightarrow du = 2ydy,{\text{ }}dy = \frac{{du}}{{2y}},{\text{ then}} \cr
& \int {\frac{{{{\left( {{y^2}} \right)}^2}y}}{{\sqrt {4 + {{\left( {{y^2}} \right)}^2}} }}} dy = \int {\frac{{{u^2}y}}{{\sqrt {4 + {u^2}} }}} \left( {\frac{1}{{2y}}} \right)du \cr
& = \frac{1}{2}\int {\frac{{{u^2}}}{{\sqrt {{{\left( 2 \right)}^2} + {u^2}} }}} du \cr
& {\text{Using the formula}} \cr
&\int {\frac{{{u^2}du}}{{\sqrt {{a^2} + {u^2}} }} = \frac{u}{2}\sqrt {{a^2} + {u^2}} - \frac{{{a^2}}}{2}\ln \left( {u + \sqrt {{a^2} + {u^2}} } \right) + C} \cr
& {\text{Therefore}}{\text{,}} \cr
& \frac{1}{2}\int {\frac{{{u^2}}}{{\sqrt {{{\left( 2 \right)}^2} + {u^2}} }}} du = \frac{u}{4}\sqrt {{{\left( 2 \right)}^2} + {u^2}} - \frac{{{{\left( 2 \right)}^2}}}{4}\ln \left( {u + \sqrt {{{\left( 2 \right)}^2} + {u^2}} } \right) + C \cr
& {\text{Simplifying}} \cr
& \frac{1}{2}\int {\frac{{{u^2}}}{{\sqrt {4 + {u^2}} }}} du = \frac{u}{4}\sqrt {4 + {u^2}} - \ln \left( {u + \sqrt {4 + {u^2}} } \right) + C \cr
& {\text{Write in terms of }}y,{\text{ }}u = {y^2},{\text{ then}} \cr
& \int {\frac{{{y^5}}}{{\sqrt {4 + {y^4}} }}} dy = \frac{{{y^2}}}{4}\sqrt {4 + {y^4}} - \ln \left( {{y^2} + \sqrt {4 + {y^4}} } \right) + C \cr} $$
