# 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 into partial fractions?1 - Factor completely polynomial Q(x) in denominator into factors of the form

__Example__

Let

The denominator is factored as follows

The quadratic term x

^{2}+ 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 1Decompose into partial fractions
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 As an exercise, group terms on the right to obtain the left side
## Example 2Decompose into partial fractions
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
## Example 3Decompose into partial fractions
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
## ExercisesDecompose 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 Exercises1. $\dfrac{3}{x-1}-\dfrac{4}{x+2}$2. $\dfrac{2}{x-3}+\dfrac{3}{(x-3)^2}$ 3. $-\dfrac{2}{x+4}+\dfrac{2 x-1}{x^2+5x+10}$ |