Find an inverse Laplace Transform using partial fractions

Question

Problem:
Find the $\mathscr{L}^{-1}\{H(s)\}$ of the following function: $$ H(s) = \dfrac{1}{s^2+5s+6} $$
Answer:
First we use partial fractions to rewrite the fraction. \begin{align*} \dfrac{1}{s^2+5s+6} &= \dfrac{A}{s+3} + \dfrac{B}{s+2} \\ 1 &= A(s+2) + B(s+3) \\ \end{align*} If we set $s = -2$ we have: $1 = A(-3+2)$. This gives us: $A = -1$. We also know that: $$ A + B = 0$$ Hence we have $B = 1$. \begin{align*} \dfrac{1}{s^2+5s+6} &= \dfrac{1}{s+2} - \dfrac{1}{s+3} \\ f(t) &= e^{-2t} - e^{-3t} \\ \end{align*} However, the book gets: $$ 2 + 5e^{5t} $$ I would like to think that I am right. Am I? I looked up the answer in the book incorrectly. The book gets: $$ e^{-2t} - e^{-3t} $$ which matches my answer.