# Partial Fraction Decomposition

How to integrate rational functions.

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## Study *Partial Fraction Decomposition* #

We know by partial fraction decomposition that every rational function \(f(x)\) possesses a representation as the sum of some polynomial \(r\) and a strict proper rational function \(\frac{p(x)}{q(x)}\), i.e \[f(x) = r(x) + \frac{p(x)}{q(x)},\quad \deg(p)<\deg(q).\] By using linearity of the integral, we can integrate each summand separately. Thus we restrict our discussion here to strict proper rational functions. Furthermore by partial fraction decomposition we can write every strict proper rational function as a sum of partial fractions \(\frac{A}{(x-x_{0})^{s}}\), with \(x_{0}\) being a root of \(q\) and \(A\) being some constant and \(s\geq 1\).

Altogether we have that it is enough to know the antiderivative of
the partial fractions in order to integrate arbitrary rational
functions.

Let us first restrict to the case \(x_{0}\in\mathbb{R}\) and \(s=1\). Then we have \[\begin{aligned}
\int \frac{A}{x-x_{0}}\, dx = A \log|x-x_{0}|+\text{const}.,\qquad
x_{0}\in\mathbb{R}.
\end{aligned}\] It is clear that this equality is only valid in
the case \(x_{0}\in\mathbb{R}\) since
the antiderivative is a real valued function.

Now consider \(x_{0}\in\mathbb{C}\setminus\mathbb{R}\).
Because of \(q\) having only real
coefficients we know that also \(\overline{x_{0}}\) is a root of \(q\) thus the partial fraction decomposition
at least has the terms \(\frac{A}{x-x_{0}}\) and \(\frac{B}{x-\overline{x_{0}}}\) included in
the sum. Then setting \(x_{0}:=\alpha+i\beta\) \[\begin{aligned}
\int \frac{A}{x-x_{0}}+\frac{B}{x-\overline{x_{0}}}\, dx &= \int
\frac{\overbrace{(A+B)}^{=:a}(x-\alpha)+i\overbrace{(A-B)}^{=:b}\beta}{\beta^{2}+(x-\alpha)^{2}}\,
dx.
\end{aligned}\] Two integrals arise in this case. For the first
one the antiderivative can be calculated using the substitution \(t=\frac{x-\alpha}{\beta}\), \(dt = \frac{dx}{\beta}\) \[\begin{aligned}
\int \frac{a(x-\alpha)}{\beta^{2}+(x-\alpha)^{2}}\, dx &= a\int
\frac{\frac{x-\alpha}{\beta}}{1+(\frac{x-\alpha}{\beta})^{2}}\,
\frac{dx}{\beta}=a\int \frac{t}{1+t^{2}}\,dt\\
& = \frac{a}{2}\log(1+t^{2})+\text{const}. =
\frac{a}2\log\left(1+\left(\frac{x-\alpha}{\beta}\right)^{2}\right)+\text{const}.
\end{aligned}\] For the second term we have again using the same
substitution as above \[\begin{aligned}
\int \frac{b\beta}{\beta^{2}+(x-\alpha)^{2}}\, dx & = b\int
\frac{1}{1+(\frac{x-\alpha}{\beta})^{2}}\, \frac{dx}{\beta}=b\int
\frac{1}{1+t^{2}}\,dt\\
& = b \arctan(t) + \text{const} = b
\arctan\left(\frac{x-\alpha}{\beta}\right) + \text{const}.
\end{aligned}\] Altogether we have for \(x_{0}\in\mathbb{C}\setminus\mathbb{R}\) and
\(s=1\) \[\begin{aligned}
\int \frac{A}{x-x_{0}}+\frac{B}{x-\overline{x_{0}}}\, dx =
\frac{A+B}2\log\left(1+\left(\frac{x-\alpha}{\beta}\right)^{2}\right)+i(A-B)\arctan\left(\frac{x-\alpha}{\beta}\right)
+ \text{const}.
\end{aligned}\] Now we want to investigate the case \(s>1\). Here it is not relevant if \(x_{0}\) is real or complex. We have \[\begin{aligned}
\int \frac{A}{(x-x_{0})^{s}}\, dx =
\frac{A}{1-s}(x-x_{0})^{1-s}+\text{const}.
\end{aligned}\]

**Example 1**. We want to integrate \(f(x)=\frac{x^{5}+2x^{3}+4x^{2}-3x}{(x^{2}+1)^{2}}\).
First of all we decompose \(f\) into a
polynomial part and a strict proper function using polynomial division
\[\frac{x^{5}+2x^{3}+4x^{2}-3x}{(x^{2}+1)^{2}} = x
+ \frac{4x^{2}-4x}{(x^{2}+1)^{2}} = x
+ \frac{4x^{2}-4x}{(x+i)^{2}(x-i)^{2}}.\] The polynomial part
has the antiderivative \[\int x\,dx =
\frac{x^2}{2}+\text{const}.\] The strict proper part has a
partial fraction decomposition \[\frac{4x^{2}-4x}{(x+i)^{2}(x-i)^{2}} =
\frac{A}{x+i}+\frac{B}{(x+i)^{2}}+\frac{\overline{A}}{x-i}+\frac{\overline{B}}{(x-i)^{2}},\]
with \(A=i\) and \(B=1-i\). Thus we have \[\begin{aligned}
\int \frac{i}{x+i} - \frac{i}{x-i}\, dx &=
2\arctan(x)+\text{const}.\\
\int \frac{1-i}{(x+i)^{2}}\,dx &= -\frac{1-i}{x+i}+\text{const}.\\
\int \frac{1+i}{(x-i)^{2}}\,dx &= -\frac{1+i}{x-i}+\text{const}.
\end{aligned}\] Altogether the antiderivative is \[\begin{aligned}
\int f(x)\, dx &= \frac12 x^2+2\arctan(x)
-\frac{1-i}{x+i}-\frac{1+i}{x-i} + \text{const}.\\
&= \frac12 x^2+2\arctan(x)-\frac{2x-2}{x^{2}+1} + \text{const}.
\end{aligned}\]

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