Answer
$-\displaystyle \frac{\sqrt{2y^{2}-3}}{y}+\sqrt{2}\ln|\sqrt{2}y+\sqrt{2y^{2}-3}|+C$
Work Step by Step
We will use the formula
42. $\displaystyle \int\frac{\sqrt{u^{2}-\mathrm{a}^{2}}}{u^{2}}du=-\frac{\sqrt{u^{2}-\mathrm{a}^{2}}}{u}+\ln|u+\sqrt{u^{2}-a^{2}}|+C$
----
$\displaystyle \int\frac{\sqrt{2y^{2}-3}}{y^{2}}dy =\qquad \left[\begin{array}{ll}
u=\sqrt{2}y , & du=\sqrt{2}dy\\
y=\frac{u}{\sqrt{2}} & dy=\frac{du}{\sqrt{2}} \\
a=\sqrt{3} &
\end{array}\right]$
$=\displaystyle \int\frac{\sqrt{u^{2}-a^{2}}}{\frac{u^{2}}{2}}\frac{du}{\sqrt{2}}=\sqrt{2}\int\frac{\sqrt{u^{2}-a^{2}}}{u^{2}}du=\qquad $...use the formula.
$=\displaystyle \sqrt{2}(-\frac{\sqrt{u^{2}-a^{2}}}{u}+\ln|u+\sqrt{u^{2}-a^{2}}|)+C$
...bring back y
$=\displaystyle \sqrt{2}(-\frac{\sqrt{2y^{2}-3}}{\sqrt{2}y}+\ln|\sqrt{2}y+\sqrt{2y^{2}-3}|)+C$
...simplify
$=-\displaystyle \frac{\sqrt{2y^{2}-3}}{y}+\sqrt{2}\ln|\sqrt{2}y+\sqrt{2y^{2}-3}|+C$
