## Calculus (3rd Edition)

$$A=1$$ $$B=-1$$ $$C=0$$
$\frac{2x+4}{(x-2)(x^2+4)}$ = $\frac{A}{x-2}$+$\frac{Bx+C}{x^2+4}$ Multiplying $(x-2)(x^2+4)$ on both sides: $2x+4=A(x^2+4)+(Bx+C)(x-2)$ If we let $x=2$ on both sides, we can eliminate constants B and C to find constant A: $4+4=A(4+4)+(Bx+C)(2-2)$ Thus, $8=8A$ and so $A=1$. Collecting all the constant terms on both sides: $4=4A-2C$ Substituting $A=1$, we find that $C=0$. Similarly, collecting all the $x$ terms on both sides: $2x=-2Bx+Cx$ If we compare only the coefficients of the $x$ terms, we arrive at the equation: $2=-2B+C$ Substituting $C=0$, we find that $B=-1$.