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Partial Fractions

Important Steps to Resolve Partial Fractions

  1. Ensure proper fraction:
    If degree(P) ≥ degree(Q), perform long division first.
  2. Factorize denominator completely.
  3. Assign constants (A, B, C, …) according to factor types.
  4. Multiply both sides by the denominator to eliminate fractions.
  5. Solve for constants by:
    • Substitution (plugging suitable x values), or
    • Comparing coefficients.
Partial Fractions – Lecture Notes | SaitechAI

Partial Fractions — Lecture Notes

SaitechAI Mathematics Lecture Series (Class 11–12, JEE/NEET Level)


1. Definition

A partial fraction expresses a rational function as a sum of simpler fractions. If \( \frac{P(x)}{Q(x)} \) is a rational function and \( \deg P(x) < \deg Q(x) \), it can be written as a sum of partial fractions.

2. Basic Rule

If \( \frac{P(x)}{Q(x)} \) is proper, its decomposition depends on the factors of \( Q(x) \):

  • Distinct Linear Factors: \( \frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b} \)
  • Repeated Linear Factors: \( \frac{P(x)}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2} \)
  • Irreducible Quadratic Factor: \( \frac{P(x)}{(x^2+bx+c)} = \frac{Ax + B}{x^2 + bx + c} \)

3. Steps to Resolve Partial Fractions

  1. Ensure the fraction is proper; if not, divide first.
  2. Factorize the denominator completely.
  3. Assign constants \(A, B, C, \dots\) based on factor types.
  4. Multiply by the denominator and remove fractions.
  5. Solve for constants by substitution or comparing coefficients.

4. Examples

Example 1:
\( \frac{3x – 5}{(x – 1)(x + 2)} = \frac{A}{x – 1} + \frac{B}{x + 2} \)
Multiply both sides by \( (x – 1)(x + 2) \): \( 3x – 5 = A(x + 2) + B(x – 1) \)
Let \( x = 1 \Rightarrow A = -\frac{2}{3} \); \( x = -2 \Rightarrow B = \frac{11}{3} \)
Final form: \( \frac{3x – 5}{(x – 1)(x + 2)} = \frac{-2/3}{x – 1} + \frac{11/3}{x + 2} \)

Example 2 (Repeated Factor):
\( \frac{2x + 3}{(x + 1)^2} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} \)
Multiply: \( 2x + 3 = A(x + 1) + B \)
Let \( x = -1 \Rightarrow B = 1 \); comparing coefficients → \( A = 2 \)
So, \( \frac{2x + 3}{(x + 1)^2} = \frac{2}{x + 1} + \frac{1}{(x + 1)^2} \)

Example 3 (Irreducible Quadratic):
\( \frac{2x^2 + 3x + 4}{(x + 1)(x^2 + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1} \)

5. Applications

  • Integration of rational functions
  • Laplace transforms
  • Electrical circuit analysis (RC, RL)
  • Control systems and differential equations

6. Common Mistakes

  • Not dividing when numerator degree ≥ denominator degree
  • Missing terms for repeated or quadratic factors
  • Incorrect coefficient comparison

7. Quick Practice Problems

  1. \( \frac{x + 2}{x^2 – 1} \)
  2. \( \frac{2x + 3}{(x – 1)^2} \)
  3. \( \frac{3x^2 + 5x + 2}{x(x + 1)(x + 2)} \)
  4. \( \frac{2x + 1}{x^2 + 4x + 5} \)

8. Integration via Partial Fractions

After decomposition, integrate each term separately:

\( \int \frac{A}{x – a} dx = A \ln|x – a| + C \)

\( \int \frac{Bx + C}{x^2 + px + q} dx = \text{Use substitution or arctan form} \)

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Worksheet – Set-1

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Problem 10 to 12

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