As first step we may use the common denominator $(x^{2}-4)=(x-2)(x+2)$ because the $\text{lcm}\left( (x-2)(x+2),(x+2)\right) =(x-2)(x+2)$ $ \begin{eqnarray*} \frac{x^{2}}{(x^{2}-4)}-\frac{x+1}{x+2} &=&\frac{x^{2}}{(x-2)(x+2)}-\frac{ \left( x+1\right) (x-2)}{\left( x+2\right) (x-2)} \\ &=&\frac{x^{2}-\left( x+1\right) (x-2)}{(x-2)(x+2)}.\tag{1} \end{eqnarray*} $
Otherwise we would get the equivalent but more more complex fraction $ \frac{x^{2}}{(x^{2}-4)}-\frac{x+1}{x+2}=\frac{x^{2}\left( x+2\right) -\left( x+1\right) (x^{2}-4)}{(x^{2}-4)\left( x+2\right) }. $ Expanding the second term of the numerator of $(1)$ $ \begin{eqnarray*} \left( x+1\right) (x-2) &=&x(x-2)+(x-2)=x^{2}-2x+x-2 \\ &=&x^{2}-x-2 \end{eqnarray*} $
and substituting into the fraction we get $ \frac{x^{2}-\left( x^{2}-x-2\right) }{(x-2)(x+2)}=\frac{x^{2}-x^{2}+x+2}{ (x-2)(x+2)}=\frac{x+2}{(x-2)(x+2)},\tag{2} $ which for $x+2\neq 0$ simplifies to $ \frac{1}{x-2}\tag{3} $
Added: In general we transform the sum (or difference) of two given rational fractions (the numerator and denominator consists of polynomials) into a single equivalent fraction, by using properties such as
- $\frac{A(x)}{B(x)}=\frac{A(x)P(x)}{B(x)P(x)}\qquad\text{for }P(x)\neq 0.$
- $\frac{A(x)}{B(x)}\pm \frac{C(x)}{D(x)}=\frac{A(x)D(x)\pm B(x)C(x)}{B(x)D(x)}.$
- $\frac{A(x)}{B_1(x)B_2(x)}\pm \frac{C(x)}{B_2(x)}=\frac{A(x)\pm B_1(x)C(x)}{B_1(x)B_2(x)}.$