# 11. Integration by Partial Fractions

by M. Bourne

If the integrand (the expression after the integral sign) is in the form of an algebraic fraction and the integral cannot be evaluated by simple methods, the fraction needs to be expressed in partial fractions before integration takes place.

The steps needed to decompose an algebraic fraction into its partial fractions results from a consideration of the reverse process − addition (or subtraction).

Consider the following addition of algebraic fractions:

1/(x+2)+5/(x+3)

=((x+3)+5(x+2))/((x+2)(x+3))

=(6x+13)/(x^2+5x+6)

In this section, we want to go the other way around. That is, if we were to start with the expression

(6x+13)/(x^2+5x+6),

and try to find the fractions whose sum gives this result, then the two fractions obtained, i.e.

1/(x+2) and 5/(x+3),

are called the partial fractions of (6x+13)/(x^2+5x+6).

We decompose fractions into partial fractions like this because:

• It makes certain integrals much easier to do, and
• It is used in the Laplace transform, which we meet later.

So if we needed to integrate this fraction, we could simplify our integral in the following way:

int(6x+13)/(x^2+5x+6)dx

=int1/(x+2)dx+int5/(x+3)dx

This is now easy to integrate:

int(6x+13)/(x^2+5x+6)dx

=int1/(x+2)dx+int5/(x+3)dx

=ln(x+2)+5\ ln(x+3)+K

=ln(x+2)(x+3)^5+K

Now we'll see how to split such a fraction into its partial fractions.

## Expressing a Fractional Function In Partial Fractions

### RULE 1:

Before a fractional function can be expressed directly in partial fractions, the numerator must be of at least one degree less than the denominator.

### Example 1

The fraction

(2x^2+3)/(x^3-1)

can be expressed in partial fractions whereas the fraction

(2x^3+3)/(x^3-1)

cannot be expressed directly in partial fractions.

However, by division

(2x^3+3)/(x^3-1) =2+5/(x^3-1)

and the resulting fraction can be expressed as a sum of partial fractions.

(Note: The denominator of the fraction must be factored before you can proceed.)

### RULE 2: Denominator Containing Linear Factors

For each linear factor (ax + b) in the denominator of a rational fraction, there is a partial fraction of the form

A/(ax+b)

where A is a constant.

### Example 2

Express the following in partial fractions. (3x)/((2x+1)(x+4)

### RULE 3: Denominator Containing Repeated Linear Factors

If a linear factor is repeated n times in the denominator, there will be n corresponding partial fractions with degree 1 to n.

For example, the partial fractions for

(5x^3+7x-9)/((x+1)^4)

will be of the form:

(5x^3+7x-9)/((x+1)^4) -=A/(x+1)+B/((x+1)^2)+C/((x+1)^3)+D/((x+1)^4)

(The sign -= means "is identically equal to". We normally apply this between 2 expressions when we wish them to be equivalent. It's OK to use the ordinary equals sign, too.)

### Example 3

(a) Express the following as a sum of partial fractions.

(x+5)/((x+3)^2)

(b) Express the following as a sum of partial fractions.

(2x^2-3)/((x-1)^3(x+1))

NOTE: Scientific Notebook can do all this directly for us using Polynomials/Partial Fractions.

### RULE 4:Denominator Containing a Quadratic Factor

Corresponding to any quadratic factor (ax^2+ bx + c) in the denominator, there will be a partial fraction of the form

(Ax+B)/(ax^2+bx+c)

### Example 4

Express the following in partial fractions.

(x^3-2)/(x^4-1)

Note: Repeated quadratic factors in the denominator are dealt with in a similar way to repeated linear factors.

Example:

(x^2+1)/((x^2+x+1)^2) -=(Ax+B)/(x^2+x+1)+(Cx+D)/((x^2+x+1)^2

## Summary

 Denominator containing… Expression Form of Partial Fractions a. Linear factor (f(x))/((x+a)(x+b)) A/(x+a)+B/(x+b) b. Repeated linear factors (f(x))/((x+a)^3) A/(x+a)+B/((x+a)^2)+C/(x+a)^3 c. Quadratic term (which cannot be factored) (f(x))/((ax^2+bx+c)(gx+h)) (Ax+B)/(ax^2+bx+c)+C/(gx+h)

Note: In each of the above cases f(x) must be of less degree than the relevant denominator.

### Exercises

Write the following fractions as sum of partial fractions and then integrate with respect to x.

1. 1/(x^2(x-2))

2. x^2/((2x+1)(x+2)^2

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