Simplify
x+1
---------
1-x^-2
Would I rationalise the denominator and multiply the top by 1+x^2
or multiply the top by x^2 and bottom I'm not sure
\(\displaystyle x \ne 0\) is implied by the original expression, so it can be multiplied by \(\displaystyle \dfrac{x^2}{x^2}\).
Moreover \(\displaystyle x\ne \pm 1.\)
\(\displaystyle \dfrac{x + 1}{1 - x^{-2}} = \dfrac{x + 1}{1 - \dfrac{1}{x^2}} = \dfrac{x + 1}{1 - \dfrac{1}{x^2}} * \dfrac{x^2}{x^2} =\\
\dfrac{x^2(x + 1)}{x^2-1} = \dfrac{x^2(x+1)}{(x + 1)(x-1)} = \dfrac{x^2}{x-1} \text { if } x \ne 0 \text { and } x \ne \pm 1.\)
You may of course multiply by \(\displaystyle \dfrac{1 +x^2}{1+x^2}\), but what good does that do?
\(\displaystyle \dfrac{x+1}{1-x{-2}} = \dfrac{x +1}{1-x^{-2}} * \dfrac{1 +x^2}{1+x^2} = \dfrac{x^3+x^2+x+1}{1 +x^2-x^{-2} - x^0} =\\
\dfrac{x^3 +x^2+x+1}{x^2-x^{-2}} \text { if } x \ne 0 \text { and } x\ne\pm1.\)
Now it is true that can be simplified further.
\(\displaystyle x \ne 1 \implies \dfrac{x^3 +x^2+x+1}{x^2 -x^{-2}} = \dfrac{\dfrac{x^4-1}{x-1}}{\dfrac{x^4-1}{x^2}} = \dfrac{x^4-1}{x-1}*\dfrac{x^2}{x^4-1} =\\
\dfrac{x^2}{x-1} \text { if } x \ne 0 \text { and } x \ne \pm 1.\)
But that is a very roundabout approach.