Partial Fractions Decompositions

Tutorial on decomposing complicated fractions into simpler manageable fractions. One of its important applications is in Integration Using Partial Fractions in calculus.

Rules of Decomposition into Partial Fractions


How to decompose a rational function rational function into partial fractions?
1 - Factor completely polynomial Q(x) in denominator into factors of the form
factored form of denominator

Example
Let
example of rational function
The denominator is factored as follows
denominator of rational function in factored form
The quadratic term x2 + 2 x + 4 is irreducible (cannot be factored) over the real.
2 - For each factor of the form $(ax + b)^m$, the decomposition includes the following sum of fractions
$\dfrac{C1}{ax + b}+\dfrac{C2}{(ax + b)^2}+...+\dfrac{Cm}{(ax + b)^m}$

Example
The fraction $\dfrac{2}{(x-2)^3}$ is decomposed as
$\dfrac{2}{(x-2)^3}=\dfrac{C1}{x-2}+\dfrac{C2}{(x-2)^2}+\dfrac{C3}{(x-2)^3}$
3 - For each factor of the form $(a x^2 + b x + c)^n$, the decomposition includes the following sum of fractions
$\dfrac{A1 x + B1}{a x^2 + b x + c} + \dfrac{A2 x + B2}{(a x^2 + b x + c)^2} + ... + \dfrac{An x + Bn}{a x^2 + b x + c)^n}$

Examples with Detailed Solutions

Example 1

Decompose into partial fractions
$\dfrac{2 x + 5}{x^2-x-2}$

Solution to Example 1:
We start by factoring the denominator
$x^2 - x - 2 = (x - 2)(x + 1)$
Both factors are linear, hence the given fraction is decomposed as follows

$\dfrac{2 x + 5}{x^2-x-2}=\dfrac{A}{x-2}+\dfrac{B}{x+1}$

Multiply both side of the above equation by the least common denominator, $(x - 2)(x + 1)$, and simplify to obtain an equation of the form
$2 x+5 = A(x + 1) + B(x - 2)$
Expand the right side and group like terms
$2 x + 5 = x (A + B) + A - 2 B$
For the right and left polynomials to be equal we need to have
$2 = A + B$ and $5 = A - 2 B$
Solve the above system to obtain
$A = 3$ and $B = -1$
Substitute $A$ and $B$ in the suggested decomposition above to obtain
$\dfrac{2 x + 5}{x^2-x-2}=\dfrac{3}{x-2}-\dfrac{1}{x+1}$

As an exercise, group terms on the right to obtain the left side

Example 2

Decompose into partial fractions
$\dfrac{1-2 x}{x^2+2x+1}$

Solution to Example 2:
We start by factoring the denominator
$x^2 + 2 x + 1 = (x + 1)^2$
Using the rule above, the given fraction is decomposed as follows

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

Multiply both side of the above equation by $(x + 1)^2$, and simplify to obtain an equation of the form
$1 - 2 x = A(x + 1) + B$
Expand the right side and group like terms
$-2x + 1 = A x + (A + B)$
For the right and left polynomials to be equal we need to have
$- 2 = A$ and $1 = A + B$
Solve the above system to obtain
$A = - 2$ and $B = 3$
Substitute $A$ and $B$ in the suggested decomposition above to obtain
$\dfrac{1-2 x}{x^2+2x+1}=\dfrac{-2}{x+1}+\dfrac{3}{(x+1)^2}$

Example 3

Decompose into partial fractions
$\dfrac{4x^2-x+8}{(x-2)(x^2+2x+3)}$

Solution to Example 3:
Use the rule above to decomposed the given fraction as follows

$\dfrac{4x^2-x+8}{(x-2)(x^2+2x+3)}=\dfrac{A}{x-2}+\dfrac{B x+C}{x^2+2x+3}$

Multiply both side of the above equation by $(x - 2)(x^2 + 2 x + 3)$, and simplify to obtain an equation of the form
$4 x^2 - x + 8 = A(x^2 + 2 x + 3) + (B x + C)(x - 2)$
The above equality is true for all values of $x$, let us use $x = 2$ to obtain an equation in $A$
$22 = 11 A$
Solve for $A$ to obtain
$A = 2$
In order to find $C$, we use $x = 0$ in the above equality
$8 = 6 - 2 C$
Solve for $C$ to obtain
$C = -1$
To find $B$, we now use $x = 1$ in the above equality
$11 = 12 + (B - 1)(1 - 2)$
Solve for $B$ to obtain
$B = 2$
The given fraction can be decomposed as follows
$\dfrac{4x^2-x+8}{(x-2)(x^2+2x+3)}=\dfrac{2}{x-2}+\dfrac{2 x-1}{x^2+2x+3}$

Exercises

Decompose the following fractions into partial fractions.
1. $\dfrac{-x+10}{x^2+x-2}$

2. $\dfrac{2 x - 3}{x-3}$

3. $\dfrac{-3 x - 24}{(x+4)(x^2+5x+10)}$

Answers to Above Exercises

1. $\dfrac{3}{x-1}-\dfrac{4}{x+2}$

2. $\dfrac{2}{x-3}+\dfrac{3}{(x-3)^2}$

3. $-\dfrac{2}{x+4}+\dfrac{2 x-1}{x^2+5x+10}$