AP Calculus BC - Area integration question

I'm taking the AP Calculus BC Exam next week and ran into this problem with no idea how to solve it. Unfortunately, the answer key didn't provide explanations, and I'd really, really appreciate it if someone could explain how to solve this problem, and why the answer is (4ln2)/5. It's a non-calculator problem.

The region is bounded by the curve $$f(x) = \frac{1}{(x+1)(4-x)}$$ and the $x$-axis from $x = 0$ to $x = 3$. What is the area of this region?

I can't seem to get the correct antiderivative of the curve.

Answer

The idea here is to use partial fraction decomposition. We set

$$\frac{1}{(x + 1)(4 - x)} = \frac{A}{x + 1} + \frac{B}{4 - x}$$
Since
\begin{align*}
\frac{A}{x + 1} + \frac{B}{4 - x} & = \frac{A(4 - x)}{(x + 1)(4 - x)} + \frac{B(x + 1)}{(x + 1)(4 - x)}\\
& = \frac{4A - Ax + Bx + B}{(x + 1)(4 - x)}\\
& = \frac{(B - A)x + 4A + B}{(x + 1)(4 - x)}
\end{align*}
we obtain
$$\frac{1}{(x + 1)(4 - x)} = \frac{(B - A)x + 4A + B}{(x + 1)(4 - x)}$$
which is an algebraic identity that holds for every real number except $x = -1$ and $x = 4$. Since the denominators are the same, we may equate the numerators, which yields

$$(B - A)x + 4A + B = 1$$
which is an algebraic identity that holds for every real number except $x = -1$ and $x = 4$. In particular, it holds when $x = 0$ and $x = 1$. Substituting $0$ for $x$ yields
$$4A + B = 1$$
Substituting $1$ for $x$ yields
$$3A + 2B = 1$$
This gives us the system of equations
\begin{alignat*}{3}
4A & + & B & = 1\\
3A & + & 2B & = 1
\end{alignat*}

Solving the system of equations yields $A = 1/5$ and $B = 1/5$. Hence,
$$\int_{0}^{3} \frac{1}{(x + 1)(4 - x)} = \frac{1}{5}\int_{0}^{3} \left(\frac{1}{x + 1} + \frac{1}{4 - x}\right) dx$$
To make the integration easier, factor out $-1$ from the numerator and denominator of $\frac{1}{4 - x}$ in order to obtain
$$\int_{0}^{3} \frac{1}{(x + 1)(4 - x)} = \frac{1}{5}\int_{0}^{3} \left(\frac{1}{x + 1} - \frac{1}{x - 4}\right) dx$$