## Calculus: Early Transcendentals 8th Edition

$-\displaystyle \frac{\sqrt{2y^{2}-3}}{y}+\sqrt{2}\ln|\sqrt{2}y+\sqrt{2y^{2}-3}|+C$
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 $=-\displaystyle \frac{\sqrt{2y^{2}-3}}{y}+\sqrt{2}\ln|\sqrt{2}y+\sqrt{2y^{2}-3}|+C$