Partial Fractions

Partial fractions are the fractions used for the decomposition of a rational expression. When an algebraic expression is split into a sum of two or more rational expressions, then each part is called a partial fraction. Hence, basically, it is the reverse of the addition of rational expressions. Similar to fractions, a partial fraction will have a numerator and denominator, where the denominator represents the decomposed part of a rational function.

In mathematics, we can see many complex rational expressions. If we try to solve the problems in a complex form, it will take a lot of time to find the solution. To avoid this complexity, we have to continue the problem by reducing the complex form of the rational expression into the simpler form. Partial fraction decomposition is one of the methods, which is used to decompose rational expressions into simpler partial fractions. This process is more useful in the integration process. In this article, you will learn the definition of the partial fraction, partial fraction decomposition, partial fractions of an improper fraction with solved examples in detail.

What is a Partial Fraction?

An algebraic fraction can be broken down into simpler parts known as “partial fractions“. Consider an algebraic fraction, (3x+5)/(2x2-5x-3). This expression can be split into simple form like [2/(x – 3)] – [1/(2x + 1)].

The simpler parts [2/(x – 3)] and [1/(2x + 1)] are known as partial fractions.

This means that the algebraic expression can be written in the form, as given in the figure:

Note: The partial fraction decomposition only works for the proper rational expression (the degree of the numerator is less than the degree of the denominator). In case, if the rational expression is an improper rational expression (the degree of the numerator is greater than the degree of the denominator), first do the division operation to convert it into proper rational expression. This can be achieved with the help of a polynomial long division method.

Partial Fraction Formulas

Here the list of Partial fractions formulas is given. These formulas will help us to decompose a rational expression into partial fractions. These are common types of partial fractions which are used to solve problems.

S.No	Rational Fraction	Partial Fraction Form
1	$\begin{array}{l}\frac{�(�)+�}{(�-�)(�-�)}\end{array}$	$\begin{array}{l}\frac{�}{�-�}+\frac{�}{(�-�)}\end{array}$
2	$\begin{array}{l}\frac{�(�)+�}{(�-�{)}^2}\end{array}$	$\begin{array}{l}\frac{{�}_1}{�-�}+\frac{{�}_2}{(�-�{)}^2}\end{array}$
3	$\begin{array}{l}\frac{�{�}^2+��+�}{(�-�)(�-�)(�-�)}\end{array}$	$\begin{array}{l}\frac{�}{�-�}+\frac{�}{(�-�)}+\frac{�}{(�-�)}\end{array}$
4	$\begin{array}{l}\frac{�{�}^2+�(�)+�}{(�-�{)}^2(�-�)}\end{array}$	$\begin{array}{l}\frac{{�}_1}{�-�}+\frac{{�}_2}{(�-�{)}^2}+\frac{�}{(�-�)}\end{array}$
5	$\begin{array}{l}\frac{�{�}^2+��+�}{(�-�)({�}^2+��+�)}\end{array}$	$\begin{array}{l}\frac{�}{�-�}+\frac{��+�}{{�}^2+��+�}\end{array}$

Here A, B and C are real numbers.

Partial Fractions of Rational Functions

Any number which can be easily represented in the form of p/q, such that p and q are integers and q≠0 is known as a rational number. Similarly, we can define a rational function as the ratio of two polynomial functions P(x) and Q(x), where P and Q are polynomials in x and Q(x)≠0. A rational function is known as proper if the degree of P(x) is less than the degree of Q(x); otherwise, it is known as an improper rational function. With the help of the long division process, we can reduce improper rational functions to proper rational functions. Therefore, if P(x)/Q(x) is improper, then it can be expressed as:

$\begin{array}{l}\frac{�(�)}{�(�)}=�(�)+\frac{�(�)}{�(�)}\end{array}$

Here, A(x) is a polynomial in x and R(x)/Q(x) is a proper rational function.

We know that the integration of a function f(x) is given by F(x) and it is represented by:

∫f(x)dx = F(x) + C

Here R.H.S. of the equation means integral of f(x) with respect to x and C is the constant of integration.

Decomposition of Partial Fractions

In order to integrate a rational function, it is reduced to a proper rational function. The method in which the integrand is expressed as the sum of simpler rational functions is known as decomposition into partial fractions. After splitting the integrand into partial fractions, it is integrated accordingly with the help of traditional integrating techniques.

The stepwise procedure for finding the partial fraction decomposition is explained here::

• Step 1: While decomposing the rational expression into the partial fraction, begin with the proper rational expression.
• Step 2: Now, factor the denominator of the rational expression into the linear factor or in the form of irreducible quadratic factors (Note: Don’t factor the denominators into the complex numbers).
• Step 3: Write down the partial fraction for each factor obtained, with the variables in the numerators, say A and B.
• Step 4: To find the variable values of A and B, multiply the whole equation by the denominator.
• Step 5: Solve for the variables by substituting zero in the factor variable.
• Step 6: Finally, substitute the values of A and B in the partial fractions.

Hence, the expression is decomposed into partial fractions.

Partial Fraction of Improper Fraction

An algebraic fraction is improper if the degree of the numerator is greater than or equal to that of the denominator. The degree is the highest power of the polynomial. Suppose, m is the degree of the denominator and n is the degree of the numerator. Then, in addition to the partial fractions arising from factors in the denominator, we must include an additional term: this additional term is a polynomial of degree n − m.

Note: 

• A polynomial with a zero degree is K, where K is a constant
• A polynomial of degree 1 is Px + Q
• A polynomial of degree 2 is Px2+Qx+K

Partial Fraction in Integration

Let us look into an example to have a better insight into integration using partial fractions.

Example: Integrate the function

$\begin{array}{l}\frac{1}{(�-3)(�+1)}\end{array}$

with respect to x.

Solution: The given integrand can be expressed in the form of partial fraction as:

$\begin{array}{l}\frac{1}{(�-3)(�+1)}=\frac{�}{(�-3)}+\frac{�}{(�+1)}\end{array}$

To determine the value of real coefficients A and B, the above equation is rewritten as:

1= A(x+1)+B(x-3)

⇒1=x(A+B)+A-3B

Equating the coefficients of x and the constant, we have

A + B = 0

A – 3B = 1

Solving these equations simultaneously, the value of A =1/4 and B = -1/4. Substituting these values in equation 1, we have

$\begin{array}{l}\frac{1}{(�-3)(�+1)}=\frac{1}{4(�-3)}+\frac{-1}{4(�+1)}\end{array}$

Integrating with respect to x we have;

$\begin{array}{l}\int \frac{1}{(�-3)(�+1)}=\int \frac{1}{4(�-3)}+\int \frac{-1}{4(�+1)}\end{array}$

According to the properties of integration, the integral of the sum of two functions is equal to the sum of integrals of the given functions, i.e.,

∫[f(x) +g(x)]dx = ∫f(x)dx + ∫g(x)dx

Therefore,

$\begin{array}{l}=\frac{1}{4}\int \frac{1}{(�-3)}–\frac{1}{4}\int \frac{1}{(�+1)}\end{array}$

$\begin{array}{l}=\frac{1}{4}\ln |�-3|–\frac{1}{4}\ln |�+1|\end{array}$

$\begin{array}{l}=\frac{1}{4}\ln \left|\frac{�-3}{�+1}\right|\end{array}$

Partial Fractions – Solved Examples

Example 1: Write the partial fraction decomposition of the following expression.

(20x + 35)/(x + 4)2

Solution:

(20x + 35)/(x + 4)2

(20x + 35)/(x + 4)2 = [A/(x + 4)] + [B/(x + 4)2]

(20x + 35)/(x + 4)2 = [A(x + 4) + B]/ (x + 4)2

Now, equating the numerators,

20x + 35 = A(x + 4) + B

20x + 35 = Ax + 4A + B

20x + 35 = Ax + (4A + B)

By equating the coefficients,

A = 20

4A + B = 35

4(20) + B = 35

B = 35 – 80 = -45

Therefore, (20x + 35)/(x + 4)2 = [20/(x + 4)] – [45/(x + 4)2]

Example 2: Decompose the given expression into partial fractions.

(x2 + 1)/ (x3 + 3x2 + 3x + 2)

Solution:

(x2 + 1)/ (x3 + 3x2 + 3x + 2)

Using the factor theorem, x + 2 is a factor of x3 + 3x2 + 3x + 2.

Thus, x3 + 3x2 + 3x + 2 = (x + 2)(x2 + x + 1)

Now, the given expression can be written as:

(x2 + 1)/ (x3 + 3x2 + 3x + 2) = (x2 + 1)/ [(x + 2)(x2 + x + 1)]

By the method of decomposition,

(x2 + 1)/(x + 2)(x2 + x + 1) = [A/(x + 2)] + [(Bx + C)/(x2 + x + 1)]

(x2 + 1)/(x + 2)(x2 + x + 1) = [A(x2 + x + 1) + (Bx + C)(x + 2)]/ [(x + 2)(x2 + x + 1)]

= [(A + B)x2 + (A + 2B + C)x + A + 2C]/ [(x + 2)(x2 + x + 1)]

Equating the coefficients in the numerators of both LHS and RHS,

A + B = 1

A + 2B + C = 0

A + 2C = 1

Solving these equations,

A = 5/3, B = -2/3 and C = -1/3

(x2 + 1)/(x + 2)(x2 + x + 1) = [5/3(x + 2)] – [(2x + 1)/3(x2 + x + 1)]

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