INTEGRATION 07
Integration Using Partial Fractions
Partial fractions rewrite a complicated rational function as a sum of simpler fractions that are easier to integrate.
The main idea
A rational function has the form
\[ \frac{P(x)}{Q(x)}. \]If the denominator can be factorised, we may be able to rewrite the fraction as a sum of simpler terms.
1. A simple example
Consider
\[ \frac{1}{(x+1)(x+2)}. \]We try to write it as
\[ \frac{A}{x+1} + \frac{B}{x+2}. \]So
\[ \frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}. \]2. Remove the denominators
Multiply both sides by
\[ (x+1)(x+2). \]Then
\[ 1 = A(x+2)+B(x+1). \]Now we solve for \(A\) and \(B\).
3. Find the constants
Let
\[ x=-1. \]Then
\[ 1=A(1), \]so
\[ A=1. \]Now let
\[ x=-2. \]Then
\[ 1=B(-1), \]so
\[ B=-1. \]Therefore,
4. Now integrate
So
\[ \int \frac{1}{(x+1)(x+2)} \,dx \]becomes
\[ \int \left( \frac{1}{x+1} - \frac{1}{x+2} \right) dx. \]Integrate term by term:
\[ = \ln|x+1| - \ln|x+2| + C. \]Therefore,
5. Why this helps
The original fraction was difficult to integrate directly.
But after decomposition, each term has the familiar form
\[ \frac{1}{x+a}, \]whose antiderivative is
\[ \ln|x+a|. \]Partial fractions therefore transform the problem into known integrals.
6. General form with two linear factors
If
\[ Q(x)=(x-a)(x-b), \]with \(a\neq b\), we try
Then determine \(A\) and \(B\).
7. Example
Decompose
\[ \frac{5x+1}{(x+1)(x+3)}. \]Write
\[ \frac{5x+1}{(x+1)(x+3)} = \frac{A}{x+1} + \frac{B}{x+3}. \]Multiply through:
\[ 5x+1 = A(x+3)+B(x+1). \]Let
\[ x=-1. \]Then
\[ -4=2A, \]so
\[ A=-2. \]Let
\[ x=-3. \]Then
\[ -14=-2B, \]so
\[ B=7. \]Hence,
\[ \boxed{ \frac{5x+1}{(x+1)(x+3)} = -\frac{2}{x+1} + \frac{7}{x+3} } \]8. Integrate the result
Therefore,
\[ \int \frac{5x+1}{(x+1)(x+3)} \,dx \]becomes
\[ -2\int\frac{1}{x+1}\,dx + 7\int\frac{1}{x+3}\,dx. \]So
\[ \boxed{ \int \frac{5x+1}{(x+1)(x+3)} \,dx = -2\ln|x+1| + 7\ln|x+3| + C } \]9. Repeated linear factors
If the denominator contains a repeated factor such as
\[ (x-a)^2, \]we need one term for each power:
For a factor
\[ (x-a)^3, \]we would use
\[ \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3}. \]10. Example with a repeated factor
Suppose
\[ \frac{1}{x(x+1)^2}. \]The correct decomposition has the form
\[ \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2}. \]Every repeated power must be included.
11. What if the numerator is too large?
Partial fractions is normally applied when the degree of the numerator is smaller than the degree of the denominator.
For example,
\[ \frac{x^3+1}{x^2-1} \]has a numerator of higher degree.
First perform polynomial division.
Then apply partial fractions to the remaining proper rational fraction.
12. Improper and proper rational functions
A rational function is called proper when
\[ \deg P<\deg Q. \]If
\[ \deg P\ge\deg Q, \]it is improper.
Improper rational functions should first be simplified using polynomial division.
13. Irreducible quadratic factors
Sometimes a denominator contains a quadratic factor that cannot be factorised over the real numbers.
For example,
\[ x^2+1. \]For such a factor, the numerator of the partial-fraction term must be linear:
rather than simply a constant numerator.
14. The practical method
15. The central idea
Partial fractions does not create a new function. It rewrites the same rational function in a form that is easier to integrate.