as you know, rationalizing the denominator simply means, "getting rid of that pesky radical at the bottom", so we'll do so by multiplying top and bottom by the "conjugate" of the denominator,
[tex]\bf \cfrac{\sqrt{3}+1}{\sqrt{3}-1}\cdot \cfrac{\sqrt{3}+1}{\sqrt{3}+1}\implies \cfrac{(\sqrt{3}+1)(\sqrt{3}+1)}{\stackrel{\textit{difference of squares}}{(\sqrt{3}-1)(\sqrt{3}+1)}}\implies \cfrac{(\sqrt{3})^2+\sqrt{3}+\sqrt{3}+1}{(\sqrt{3})^2-1^2}
\\\\\\
\cfrac{3+2\sqrt{3}+1}{3-1}\implies \cfrac{4+2\sqrt{3}}{2}\implies \cfrac{\underline{2}(2+1\sqrt{3})}{\underline{2}}\implies 2+\sqrt{3}[/tex]
