# Partial Fraction Decomposition methods

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?

• It is not a trick but the method. – Claude Leibovici Feb 12 at 13:53

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.)

This is usual partial fraction decomposition. The referenced Wikipedia article provides algorithms to compute the decomposition.