Perform polynomial division or algebraic manipulation to make the fraction proper.

Since the degree of the numerator is equal to the degree of the denominator, we first divide the numerator by the denominator.

$$\frac{x^2+2}{x^2-4} = \frac{(x^2-4)+6}{x^2-4} = \frac{x^2-4}{x^2-4} + \frac{6}{x^2-4} = 1 + \frac{6}{x^2-4}$$

The denominator $x^2-4$ is a difference of squares:

$$x^2-4 = (x-2)(x+2)$$

Multiply both sides by $(x-2)(x+2)$:

$$6 = A(x+2) + B(x-2)$$

To find $A$, set $x=2$:

$$6 = A(2+2) + B(2-2)$$ $$6 = 4A + 0$$ $$4A = 6$$ $$A = \frac{6}{4} = \frac{3}{2}$$

To find $B$, set $x=-2$:

$$6 = A(-2+2) + B(-2-2)$$ $$6 = 0 + B(-4)$$ $$-4B = 6$$ $$B = \frac{6}{-4} = -\frac{3}{2}$$

$$\frac{x^2+2}{x^2-4} = 1 + \frac{3}{2(x-2)} - \frac{3}{2(x+2)}$$

The expression in partial fractions is $1 + \frac{3}{2(x-2)} - \frac{3}{2(x+2)}$.