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05 - Partial Fraction

Partial Fraction

  • To express a single rational fraction as a sum of two or more rational fractions is called Partial fraction.
  • Expressing a rational function as a sum of partial fraction is called Partial Fraction Resolution.
  • It plays an important part in the study of calculus

Terms Regarding Partial Fraction

  • Conditional Equation: It is an equation is which two two algebraic expressions are equal for particular values of variables.

    Example: \(2x=3\)

  • Identity: An equation that is true for all the value of \(x\) is called identity.

    Example: \((x+3)(x+4)=x^2+7x+12\)

  • Rational Fraction: The quotient of two polynomials \(\frac{P(x)}{Q(x)}\), \(Q\rlap{/}{=}0\) with no common factors, is called a rational fraction. Rational fractions are of two types.

    Example: $$ \frac{x^3 - x^2 + x+1}{x^2 + 5} $$

  • Proper Fraction: The rational fraction \(\frac{P(x)}{Q(x)}\) is called proper if degree of numerator (\(P(x)\)) is less than degree of denominator (\(Q(x)\)).

    Example: $$ \frac{2x-5}{x^2+4} $$

  • Improper Fraction: The rational fraction \(\frac{P(x)}{q(x)}\) is called proper if degree of numerator (\(P(x)\)) is greater than or equal to degree of denominator (\(Q(x)\)).

    Example: $$ \frac{x}{2x-3} $$

Partial Fraction Resolution.

Following are the main points of resolving a rational fraction \(\frac{P(x)}{Q(x)}\) into partial fraction:

  • The degree of \(P(x)\) should be less than the degree of \(Q(x)\). If not, divide and work with reminder theorem.
  • Clear the given equation of fractions.
  • Equate the coefficients of like terms (power of \(x\))
  • Solve the resulting equation for coefficients.

There are following cases of partial fraction resolution from a rational fraction $$ \frac{P(x)}{Q(x)} $$

Case 1

When \(Q(x)\) has only non-repeating linear factors

Example

\[ \frac{x-7}{(x-1)(x+2)} \]

Solution

This case can be solved easily by using cover up method:

  • As we known the above equation will be written as: $$ \frac{x-7}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2} $$
  • To find \(A\) cover \(x-1\) and put \(x=1\) on right side: $$ \frac{(1)-7}{(1)+2}=\frac{-6}{3}=-2 $$
  • To find \(B\) cover \(x+2\) and put \(x=-2\) on right side: $$ \frac{(-2)-7}{(-2)-1}=\frac{-9}{-3}=3 $$
  • So $$ \frac{x-7}{(x-1)(x+2)}=\frac{-2}{x-1}+\frac{3}{x+2} $$

Case 2

When \(Q(x)\) has repeating linear factors

Example

\[ \frac{x^2 + x-1}{(x+2)^3} \]

Solution

Checkout textbook solution.

Case 3

When \(Q(x)\) has non-repeating irreducible quadratic factors

Example

\[ \frac{3x-11}{(x^2 +1)(x+3)} \]

Solution

Checkout textbook solution.

Case 4

When \(Q(x)\) has repeating irreducible quadratic factors

Example

\[ \frac{3x-11}{(x^2 +1)^2(x+3)} \]

Solution

Checkout textbook solution.

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