## Calculus (3rd Edition)

$$B=-2$$
$\frac{3x^2+11x+12}{(x+1)(x+3)^2}$ = $\frac{1}{x+1}$-$\frac{B}{x+3}$-$\frac{3}{(x+3)^2}$ Multiplying $(x+1)(x+3)^2$ on both sides: $3x^2+11x+12 = (x+3)^2-B(x+1)(x+3)-3(x+1)$ Collecting all the $x^2$ terms on both sides of the equation, we have: $3x^2 = x^2 -Bx^2$ Comparing only the coefficients of the $x^2$ terms: $3=1-B$ and hence, $B=-2$.