Partial fraction decomposition rewrites one rational function as a sum of simpler rational functions.
It is especially useful before integrating rational functions.
1. What is a rational function?
A rational function is a quotient of two polynomials:
\[
\begin{aligned}
\frac{P(x)}{Q(x)}.
\end{aligned}
\]
Partial fractions are used when \(Q(x)\) can be factored into simpler pieces.
2. Proper and improper rational functions
A rational function is proper when:
\[
\begin{aligned}
\deg P(x) < \deg Q(x).
\end{aligned}
\]
If the numerator degree is greater than or equal to the denominator degree, polynomial long division must be done first:
\[
\begin{aligned}
\frac{P(x)}{Q(x)}
=
S(x)+\frac{R(x)}{Q(x)}.
\end{aligned}
\]
Then only the proper fraction \(R(x)/Q(x)\) is decomposed.
3. Linear factors
If the denominator has distinct linear factors, use one constant numerator for each factor:
\[
\begin{aligned}
\frac{P(x)}{(x-a)(x-b)}
=
\frac{A}{x-a}
+
\frac{B}{x-b}.
\end{aligned}
\]
The constants \(A\) and \(B\) are found by clearing denominators and matching coefficients.
4. Worked example: \((2x+3)/(x^2-1)\)
Start with:
\[
\begin{aligned}
\frac{2x+3}{x^2-1}.
\end{aligned}
\]
Factor the denominator:
\[
\begin{aligned}
x^2-1
=
(x-1)(x+1).
\end{aligned}
\]
Set up the template:
\[
\begin{aligned}
\frac{2x+3}{(x-1)(x+1)}
=
\frac{A}{x-1}
+
\frac{B}{x+1}.
\end{aligned}
\]
Multiply both sides by \((x-1)(x+1)\):
\[
\begin{aligned}
2x+3
=
A(x+1)+B(x-1).
\end{aligned}
\]
Expand and match coefficients:
\[
\begin{aligned}
2x+3
&=
(A+B)x+(A-B).
\end{aligned}
\]
Therefore:
\[
\begin{aligned}
A+B&=2,\\
A-B&=3.
\end{aligned}
\]
Solving gives:
\[
\begin{aligned}
A&=\frac{5}{2},\\
B&=-\frac{1}{2}.
\end{aligned}
\]
So:
\[
\begin{aligned}
\boxed{
\frac{2x+3}{x^2-1}
=
\frac{5}{2}\cdot\frac{1}{x-1}
-
\frac{1}{2}\cdot\frac{1}{x+1}
}.
\end{aligned}
\]
5. Cover-up method
For distinct linear factors, the cover-up method can quickly find constants.
For:
\[
\begin{aligned}
\frac{2x+3}{(x-1)(x+1)}
=
\frac{A}{x-1}
+
\frac{B}{x+1},
\end{aligned}
\]
cover \(x-1\), then substitute \(x=1\):
\[
\begin{aligned}
A
&=
\frac{2(1)+3}{1+1}
=
\frac{5}{2}.
\end{aligned}
\]
Cover \(x+1\), then substitute \(x=-1\):
\[
\begin{aligned}
B
&=
\frac{2(-1)+3}{-1-1}
=
-\frac{1}{2}.
\end{aligned}
\]
6. Repeated linear factors
If a linear factor repeats, include every power up to the repeated power:
\[
\begin{aligned}
\frac{P(x)}{(x-a)^3}
=
\frac{A_1}{x-a}
+
\frac{A_2}{(x-a)^2}
+
\frac{A_3}{(x-a)^3}.
\end{aligned}
\]
A common mistake is to include only the highest power.
All lower powers must be included.
7. Irreducible quadratic factors
If the denominator contains an irreducible quadratic factor, use a linear numerator:
\[
\begin{aligned}
\frac{P(x)}{x^2+1}
=
\frac{Bx+C}{x^2+1}.
\end{aligned}
\]
The numerator degree must be less than the denominator factor degree.
Since the denominator is quadratic, the numerator must be linear.
8. Mixed linear and quadratic factors
For example:
\[
\begin{aligned}
\frac{4x+1}{(x-1)(x^2+1)}
=
\frac{A}{x-1}
+
\frac{Bx+C}{x^2+1}.
\end{aligned}
\]
After clearing denominators, match coefficients of powers of \(x\).
9. Coefficient matching
Coefficient matching means comparing equal powers of \(x\).
If:
\[
\begin{aligned}
2x+3
=
(A+B)x+(A-B),
\end{aligned}
\]
then the coefficient of \(x\) on the left must equal the coefficient of \(x\) on the right,
and the constant term on the left must equal the constant term on the right.
10. Why partial fractions help in integration
Many rational functions are hard to integrate in one piece, but easier after decomposition:
\[
\begin{aligned}
\int \frac{2x+3}{x^2-1}\,dx
&=
\int
\left(
\frac{5}{2}\cdot\frac{1}{x-1}
-
\frac{1}{2}\cdot\frac{1}{x+1}
\right)\,dx.
\end{aligned}
\]
Each term now has a simpler denominator.
11. Formula summary
The table below uses plain text formulas in table cells to avoid raw LaTeX rendering problems.
12. Common mistakes
- Trying to decompose an improper rational function before long division.
- Forgetting lower powers for repeated linear factors.
- Using a constant numerator over an irreducible quadratic instead of \(Bx+C\).
- Not factoring the denominator completely before writing the template.
- Dropping negative signs while clearing denominators.
- Using the cover-up method when factors are repeated or quadratic.
Key idea: divide first if needed, factor the denominator, write the correct template, then solve the unknown coefficients.
