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Nvm I found the answer 

You may want to see a slick way of solving this problem.  It depends on knowing this fact.
If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$.  For your problem it goes: We first note that $\frac{\frac{x+1}{x}}{\frac{x-1}{x}} = \frac{x+1}{x-1}$.  so
\begin{eqnarray*}
\dfrac{x+1}{x-1}
&=& \frac{5}{4} \\
\dfrac{(x+1)+(x-1)}{(x+1)-(x-1)}
&=& \dfrac{5+4}{5-4} \\
\dfrac{2x}{2}
&=& \dfrac{9}{1} \\
x &=& 9
\end{eqnarray*}

OK then.   You can still use my technique for this problem.
Note that \[\dfrac{x+\dfrac{1}{x}}{x-\dfrac{1}{x}} =
\dfrac{x^2+1}{x^2-1}.\]
So
\begin{eqnarray*}
    \dfrac{x^2+1}{x^2-1} &=& \dfrac{5}{4} \\
    \dfrac{(x^2+1)+(x^2-1)}{(x^2+1)-(x^2-1)} &=& \dfrac{5+4}{5-4} \\
    \dfrac{2x^2}{2} &=& \dfrac{9}{1} \\
    x^2 &=& 9 \\
    x &=& \pm 3.
\end{eqnarray*}