0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ It is not a trick but the method. $\endgroup$ – Claude Leibovici Feb 12 at 13:53
3
$\begingroup$

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.)

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.