Ucale

# Partial Fraction

InÂ theÂ partial fraction decompositionÂ orÂ partial fraction expansionÂ of a rational number (that is, a fractionÂ such that the numerator and the denominator are both polynomialÂ Â is an operation that consists of expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.

If f(x) and g(x) are two polynomialÂ $\displaystyle \frac { f\left( x \right) }{ g\left( x \right) }Â$ defines a rational algebraic function or a rational function of x.

If degree of f(x) < degree of g(x), thenÂ $\displaystyle \frac { f\left( x \right) }{ g\left( x \right) }Â$ is called a proper rational function.

$\displaystyle If\quad degree\quad of\quad f\left( x \right) \ge degree\quad of\quad g\left( x \right) \quad then\quad \frac { f\left( x \right) }{ g\left( x \right) } Â$

IfÂ $\displaystyle \frac { f\left( x \right) }{ g\left( x \right) }Â$ is an improper rational function, we divideÂ f(x) by g(x) so that the rational functionÂ $\displaystyle \frac { f\left( x \right) }{ g\left( x \right) }Â$ is expressed in the formÂ $\displaystyle \phi \left( x \right) +\frac { \psi \left( x \right) }{ g\left( x \right) } Â$ whereÂ $\displaystyle \phi \left( x \right) \quad and\quad \psi \left( x \right) Â$ are polynomial such that the degree ofÂ $\displaystyle \psi \left( x \right)$ is less than that of g(x). Thus ,$\displaystyle \frac { f\left( x \right) }{ g\left( x \right) }Â$Â  is expressible as the sum of a polynomial and a proper rational function.

Any proper rational functionÂ  $\displaystyle \frac { f\left( x \right) }{ g\left( x \right) }Â$Â  can be expressed as the sum of rational functions, each having a simple factor of g(x) . EachÂ  such fraction is called partial fraction and the process of obtaining them is called the resolution or decomposition ofÂ $\displaystyle \frac { f\left( x \right) }{ g\left( x \right) }Â$ into partial fraction.

Any resolution ofÂ $\displaystyle \frac { f\left( x \right) }{ g\left( x \right) }Â$ into partial fractions depends mainly upon the nature of the fraction g(x) as discussed below

Case IÂ

When denominator is expressible as the product of non-repeating liner factors

$\displaystyle Let\quad g\left( x \right) =\left( x-{ a }_{ 1 } \right) \left( x-{ a }_{ 2 } \right) ......\left( x-{ a }_{ n } \right) Â$ Then we assume that

$\displaystyle \frac { f\left( x \right) }{ g\left( x \right) } =\frac { { A }_{ 1 } }{ x-{ a }_{ 1 } } +\frac { { A }_{ 2 } }{ x-{ a }_{ 2 } } +.....\frac { { A }_{ n } }{ x-{ a }_{ n } } \\ where\quad { A }_{ 1 },{ A }_{ 2 },....{ A }_{ n }$

are constants and can be determined by equating the numerator on RHS to the numerator on LHS and then substitutingÂ $\displaystyle x={ a }_{ 1 },{ a }_{ 2 },....{ a }_{ n }$

#### Example

ResolveÂ $\displaystyle \frac { 3x+2 }{ { x }^{ 3 }-6{ x }^{ 2 }+11x-6 }$ into partial fractions

Solution:

$\displaystyle we\quad have\quad \frac { 3x+2 }{ { x }^{ 3 }-6{ x }^{ 2 }+11x-6 } =\frac { 3x+2 }{ \left( x-1 \right) \left( x-2 \right) \left( x-3 \right) } \\ Let\quad \frac { 3x+2 }{ \left( x-1 \right) \left( x-2 \right) \left( x-3 \right) } =\frac { A }{ \left( x-1 \right) } +\frac { B }{ \left( x-2 \right) } +\frac { C }{ \left( x-3 \right) } \quad Then$

$\displaystyle \Rightarrow \frac { 3x+2 }{ \left( x-1 \right) \left( x-2 \right) \left( x-3 \right) } =\frac { A\left( x-2 \right) \left( x-3 \right) +B\left( x-1 \right) \left( x-3 \right) +C\left( x-1 \right) \left( x-2 \right) }{ \left( x-1 \right) \left( x-2 \right) \left( x-3 \right) } \\ \Rightarrow 3x+2=A\left( x-2 \right) \left( x-3 \right) +B\left( x-1 \right) \left( x-3 \right) +C\left( x-1 \right) \left( x-2 \right) \qquad ......\left( i \right)Â$

Putting x-1 =0 or x=1 in (i), we get 5=A(1-2)(1-3)$\displaystyle \Rightarrow A=\frac { 5 }{ 2 } Â$

Putting x-2 =0 or x=2 in (i), we get 8=B(2-1)(2-3)$\displaystyle \Rightarrow B=-8$

Putting x-3=0 or x=3 in (i), we get 11=C(3-1)(3-2)$\displaystyle \Rightarrow C=\frac { 11 }{ 2 } Â$

$\displaystyle \therefore \frac { 3x+2 }{ { x }^{ 3 }-6{ x }^{ 2 }+11x-6 } \\ =\frac { 3x+2 }{ \left( x-1 \right) \left( x-2 \right) \left( x-3 \right) } =\frac { 5 }{ 2\left( x-1 \right) } -\frac { 8 }{ \left( x-2 \right) } +\frac { 11 }{ 2\left( x-3 \right) }$

April 18, 2019
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