Taking the integral by partial fractions of $\frac{4}{(x+a)(x+b)}$ Given this equation, how do I solve for the integral if there are so many variables that I can't find $A$ and $B$?
$$\int \frac4{(x+a)(x+b)}\mathrm{d}x$$
I got until the point:
$$ \frac4{(x+a)(x+b)} = \frac{A(x+b)+B(x+a)}{(x+a)(x+b)}$$
so taking the numerator of the entire equation,
$$ 4 = {A(x+b)+B(x+a)}$$
but then I don't know what to plug in to find $A$ and $B$.

Ok, you get to $$4 = {A(x+b)+B(x+a)}.$$
Let's transform it:
$$4 = {A(x+b)+B(x+a)} \Rightarrow \\
0x + 4 = x(A+B) + (Ab + Ba).$$
Then you need to solve:
$$\begin{cases}
A+B = 0\\
Ab + Ba = 4
\end{cases} \Rightarrow \begin{cases}
A = -B\\
A(b -a) = 4
\end{cases} \Rightarrow \begin{cases}
A = \frac{4}{b-a}\\
B = \frac{4}{a-b}
\end{cases}.$$
