Taking the integral by partial fractions of $\frac{4}{(x+a)(x+b)}$

Question

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 $B$.

Answer 1

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:

$$\begin{cases} A+B = 0\\ Ab + Ba = 4 \end{cases} \Rightarrow \begin{cases} A = -B\\ A(b -a) = 4 \end{cases} \Rightarrow \begin{cases} A = \frac{4}{b-a}\\ B = \frac{4}{a-b} \end{cases}.$$