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Partial Fractions

Top
 Sub Topics A rational expression is a quotient of polynomials, that is an expression in the form $\frac{p(x)}{q(x)}$ where p(x) and q(x) are polynomials. A numerical fraction can be expressed as a sum of smaller fractions. For example, we can write,$\frac{5}{8}$ = $\frac{1}{2}$ + $\frac{1}{8}$. Same way a rational expression can also be expressed as sum of two are more fractional expressions. The technique of decomposing a rational expression into its partial fractions simplifies the process of solving complex problems.

Defintion

A rational expression $\frac{p(x)}{q(x)}$ is said to be proper if the degree of the numerator is less than that of the denominator, just as a numerical proper fraction. A proper rational function can be expanded into a sum of its partial fractions. The sum contains the same number of fractions as the number of factors in the denominator.

Suppose q(x) is factored as q(x) = $q_1(x) . q_2(x) . q_3(x)$

Then the partial fraction expansion for $\frac{p(x)}{q(x)}$ is of the form

$\frac{p(x)}{q(x)}$ = $\frac{p_{1}(x)}{q_{1}(x)}$ + $\frac{p_{2}(x)}{q_{2}(x)}$ + $\frac{p_{3}(x)}{q_{3}(x)}$

where each of the fractions in the sum are proper.

Partial Fraction Decomposition Method

The principle technique applied in the process of Partial Fraction expansion is finding the expressions for the numerators of the partial fractions. The following table gives the term/s in the partial fraction corresponding to the type of the factor in the denominator.

 Type of the factor Example Term in the partial fraction expansion Single Linear Factor (ax + b) $\frac{A}{ax+b}$ Repeated Linear Factor $(ax + b)^2$ $\frac{A_{1}}{ax+b}$ + $\frac{A_{2}}{(ax+b)^{2}}$ Single Quadratic Factor $ax^2 + bx + c$ $\frac{Ax+B}{ax^{2}+bx+c}$ Repeated Quadratic Factor $(ax^2+ bx + c)^2$ $\frac{A_{1}x+B_{1}}{ax^{2}+bx+c}$ + $\frac{A_{2}x+B_{2}}{(ax^{2}+bx+c)^{2}}$

Repetition of factors contribute to addition of terms in the expansion. Similar patterns in expansion can be used for factors of higher degrees.
If the degree of the numerator is greater than that of the denominator, first divide and separate the quotient. Then the partial decomposition can be done to the proper fraction formed with the remainder and denominator of the original fraction.

Method of Partial Fractions

The steps involved in the method of Partial Fractions can be listed as follows:
1. Factor the denominator.
2. Write the terms corresponding to the type of factors as given in the table for Partial fraction decomposition.
3. Find the values of constants used in the Partial Fraction expansion using the algebraic technique of substituting some convenient values to the variable on either side.
4. The values of constants can also be found by equating the like coefficients.

How to do Partial Fractions

Let us expand a rational expression into Partial fractions to understand the techniques used.

Decompose into partial fractions $\frac{3x^{2}-7x-2}{x^{3}-x}$.

The degree of the numerator is less than that of the denominator. Hence the the rational expression can be expanded into partial fractions.

Factoring the denominator, the given expression can be written as:

$\frac{3x^{2}-7x-2}{x^{3}-x}$ = $\frac{3x^{2}-7x-2}{x(x+1)(x-1)}$

The denominator has three distinct linear factors. Hence the decomposition is of the form:

$\frac{3x^{2}-7x-2}{x(x+1)(x-1)}$ = $\frac{A}{x}$ + $\frac{B}{x+1}$ + $\frac{C}{x-1}$

Multiplying the equation by x(x +1)(x - 1) we get the Basic equation to solve for the constants A, B and C as,

$3x^2 - 7x - 2 = A(x + 1)(x - 1) + Bx(x -1) + Cx(x + 1)$

By substituting x = 0, 1 and -1 in the equation, we get,

- 2 = -A Set x = 0

3 - 7 - 2 = 2C Set x = 1

3 + 7 -2 = 2B Set x = -1

Thus we get A = 2, B = 4 and C = -3

Hence the partial fraction decomposition is:

$\frac{3x^{2}-7x-2}{x^{3}-x}$ = $\frac{2}{x}$ + $\frac{4}{x+1}$ - $\frac{3}{x-1}$

Integration by Partial Fraction

Partial fraction decomposition is an important technique used for integrating rational functions, which leads to straightforward integration steps.

Example: Integrate $\int$ $\frac{x-1}{x^{2}+3x+2}$ $dx$

let us first do the partial fraction expansion for the integrand.

$\frac{x-1}{x^{2}+3x+2}$ =$\frac{x-1}{(x+2)(x+1)}$

$\frac{x-1}{(x+2)(x+1)}$ = $\frac{A}{x+2}$ + $\frac{B}{x+1}$

Multiplying the equation by (x + 2)(x + 1) we get the basic equation to solve for A and B as

x - 1 = A(x + 1) + B(x + 2)

-2 = B Set x = -1

-3 = -A Set x = -2

Hence A = 3 and B = -2.

The integral can be hence separated as:

$\int$ $\frac{x-1}{x^{2}+3x+2}$ $dx$ = $\int$ $\frac{3}{x+2}$ -$\int \frac{2}{x+1}$

= 3ln| x + 2 | - 2ln| x + 1 | = $ln$ $(\frac{|x+2|^{3}}{{|x+1|^{2}}})$

Examples

Example 1: Decompose into partial fractions $\frac{4x^{2}-1}{2x(x+1)^{2}}$

The denominator has one linear factor and one repeating linear factor.

Hence the decomposition format is:

$\frac{4x^{2}-1}{2x(x+1)^{2}}$ = $\frac{A}{x}$ + $\frac{B}{x+1}$ + $\frac{C}{(x+1)^{2}}$

Multiplying the equation by $x(x + 1)^2$ we get the Basic equation to solve for A, B and C as.

$4x^2 - 1 = A(x + 1)^2 + Bx(x + 1) + Cx$

-1 = A Set x = 0

4 - 1 = -C Set x = -1

4 - 1 = 4A + 2B + C Set x = 1

A = -1 , C = -3 and 4A + 2B + C = 3 ? B = 5

Hence $\frac{4x^{2}-1}{2x(x+1)^{2}}$ = -$\frac{1}{x}$ + $\frac{5}{x+1}$ - $\frac{3}{(x+1)^{2}}$

Example 2:  Expand as a sum of partial fractions $\frac{1}{x^{4}-1}$

The factored form of the expression is:

$\frac{1}{x^{4}-1}$ = $\frac{1}{(x+1)(x-1)(x^{2}+1)}$

The denominator has two linear and one quadratic factor. Hence the format for partial fraction expansion is.

$\frac{1}{(x+1)(x-1)(x^{2}+1)}$ = $\frac{A}{x+1}$ + $\frac{B}{x-1}$ +$\frac{Cx+D}{x^{2}+1}$

Multiplying the equation by $(x + 1)(x - 1)(x^2 + 1)$ we get the Basic equation to solve for constants A,B, C and D as
1 = $A(x - 1)(x^2 + 1) + B(x + 1)(x^2 + 1) + (Cx + D)(x^2 - 1)$

Ordering the right side of the equation in descending power of x.

1 = $(A + B + C)x^3+ (-A + B + D)x^2 + (A + B - C)x -A + B -D$

Equating the like coefficients on either side.

0 = A + B + C ( Coefficients of $x^3$.)

0 = -A + B + D (Coefficients of $x^2$).

0 = A + B - C  (Coefficients of x).

1 = -A + B - D (Constants)

Solving the above four equations simultaneously we get

A = -$\frac{1}{4}$ B = $\frac{1}{4}$ C= 0 and D = -$\frac{1}{2}$.

Hence the required partial fraction expansion is.

$\frac{1}{x^{4}-1}$ = -$\frac{1}{4(x+1)}$ + $\frac{1}{4(x-1)}$ -$\frac{1}{2(x^{2}+1)}$