I have this problem where I have to solve the following integral.
${\displaystyle\int}\dfrac{1}{\left(x+a^2\right)\left(x+b\right)^2}\,\mathrm{d}x$
I checked the solution on https://www.integral-calculator.com/ and the first step of the solution is a trick.
$={\displaystyle\int}\left(-\dfrac{1}{\left(b^2-2a^2b+a^4\right)\left(x+b\right)}-\dfrac{1}{\left(b-a^2\right)\left(x+b\right)^2}+\dfrac{1}{\left(b^2-2a^2b+a^4\right)\left(x+a^2\right)}\right)\mathrm{d}x$
So here the fraction is decomposed into 3 parts. Afterwards the integral is easily solved by applying linearity. How to determine in how many parts the fraction should be decomposed and how to actually decompose it without guessing?
Here $x = -b$ is a pole of order $2$ and $x = -a^2$ is a pole of order $1$. There is no quadratic or higher order terms here in the denominator.
So the partial fraction will consist of $2+1 = 3$ terms.
In general (considering only linear factors in the denominator)
$\frac{c}{(x-p_1)^{k}(x-p_2)^{h}}$ ($k$ and $h$ are positive integers) can be split into $k+h$ partial fractions as follows.
$$\begin{align}\frac{c}{(x-p_1)^{k}(x-p_2)^{h}} &= \left[\frac{l_1}{x-p_1} +\frac{l_2}{(x-p_1)^2} +\cdots+\frac{l_k}{(x-p_1)^k}\right] \\&+\left[ \frac{m_1}{x-p_2} +\frac{m_2}{(x-p_2)^2} +\cdots\frac{m_h}{(x-p_2)^h} \right]\end{align} $$
($l_i$ and $h_i$ are constants to be determined.)