simplifying complex fractions I have this expression
${2 \over x^2 - 4}$ + ${1 \over x + 2}$
So I would take the LCM to ${x^2 - 4}$ or using the difference of 2 squares to be ${(x + 2)(x - 2)}$
So I would simplify this to:
${2 + (x - 2) \over (x - 2)(x + 2)}$ which would simplify further to
${2\over x + 2}$
But the answer to the question is ${x \over x^2 - 4}$
I'm not sure how the real answer is achieved and how I am wrong
 A: I do not believe either answer is correct (although in the case you forgot to subtract the 2 from the other 2 to achieve just $x$ in the numerator). 
You would indeed multiply the numerator and denominator top by $(x-2)$, but the result should be different.
$$ \frac{2x}{x^2-4} + \frac{x-2}{x^2-4} = \frac{3x-2}{x^2-4}$$
Which does not simplify further.
EDIT: Going off Chinny's (correct) thought that you intended $\frac{2}{x^2-4} + \frac{1}{x+2}$ your error is then in thinking you could cancel out factors in $\frac{2+(x-2)}{(x+2)(x-2)}$, which you cannot. That would only work with a numerator involving multiplication or division (consider the fact that if you really were to manually divide the numerator and denominator by $(x-2)$ you'd get the cancelation, but still have to deal with the $\frac{2}{(x-2)}$ part. (Note that you could still push this through, you'd have $$\frac{\frac{2}{x-2}+1}{x+2} \implies \frac{\frac{2+(x-2)}{x-2}}{x+2} \implies \frac{x}{(x+2)(x-2)}.)$$
A: Notice, we have $$\frac{2}{x^2-4}+\frac{1}{x+2}$$ $\color{red}{\text{apply} \ a^2-b^2=(a-b)(a+b)}$ $$=\frac{2}{(x-2)(x+2)}+\frac{1}{x+2}$$ $\color{red}{\text{take L.C.M of denominators }}$  $$=\frac{2+x-2}{(x-2)(x+2)}$$$$ =\frac{x}{(x-2)(x+2)}$$ Thus we have following simplified form  $$\bbox[5px, border: 2px solid #C0A000]{\color{red}{\frac{2}{x^2-4}+\frac{1}{x+2}=\frac{x}{(x-2)(x+2)}}}$$
