Answer
$\dfrac{\sqrt{p}+\sqrt{p^{2}-1}}{\sqrt{p}-\sqrt{p^{2}-1}}=\dfrac{p^{2}+p+2\sqrt{p^{3}-p}-1}{-p^{2}+p+1}$
Work Step by Step
$\dfrac{\sqrt{p}+\sqrt{p^{2}-1}}{\sqrt{p}-\sqrt{p^{2}-1}}$
Multiply the numerator and the denominator of the given expression by $\sqrt{p}+\sqrt{p^{2}-1}$, which is the conjugate of the denominator:
$\dfrac{\sqrt{p}+\sqrt{p^{2}-1}}{\sqrt{p}-\sqrt{p^{2}-1}}\cdot\dfrac{\sqrt{p}+\sqrt{p^{2}-1}}{\sqrt{p}+\sqrt{p^{2}-1}}=...$
Evaluate the products:
$...=\dfrac{(\sqrt{p}+\sqrt{p^{2}-1})^{2}}{(\sqrt{p})^{2}-(\sqrt{p^{2}-1})^{2}}=...$
$...=\dfrac{(\sqrt{p})^{2}+2(\sqrt{p})(\sqrt{p^{2}-1})+(\sqrt{p^{2}-1})^{2}}{p-p^{2}+1}=...$
$...=\dfrac{p+2\sqrt{p^{3}-p}+p^{2}-1}{-p^{2}+p+1}=\dfrac{p^{2}+p+2\sqrt{p^{3}-p}-1}{-p^{2}+p+1}$
