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
How to decompose a fraction $\dfrac{P(x)}{Q(x)}$ into partial fractions?
1  Factor completely polynomial $Q(x)$ into factors of the form
Example
Let $f(x) = \dfrac {(2x  1)}{(x^3+ 2 x^2 + 4 x)}$
The denominator is factored as follows
$(x^3 + 2 x^2 + 4 x) = x(x^2 + 2 x + 4)$
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 $\dfrac{2}{(x2)^3}$ is decomposed as
$\dfrac{2}{(x2)^3}=\dfrac{C1}{x2}+\dfrac{C2}{(x2)^2}+\dfrac{C3}{(x2)^3}$
3  For each factor of the form $(a x^2 + b x + c)^n$, the decomposition includes the following sum of fractions
$\dfrac{A1 x + B1}{a x^2 + b x + c} + \dfrac{A2 x + B2}{(a x^2 + b x + c)^2} + ... + \dfrac{An x + Bn}{a x^2 + b x + c)^n}$
Example 1: Decompose into partial fractions
Solution to Example 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 As an exercise, group terms on the right to obtain the left side
Example 2: Decompose into partial fractions
Solution to Example 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
Example 3: Decompose into partial fractions
Solution to Example 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
Exercises: Decompose the following fractions into partial fractions.
