Taming Complex Fractions: Integration by Partial Fractions
Master the systematic approach to integrate rational functions by decomposing them into simpler fractions. Essential for JEE calculus preparation.
Why Partial Fractions Matter in JEE
Integration by partial fractions appears in every JEE mathematics paper, accounting for 4-8 marks. This technique transforms complex rational functions into simpler integrals that can be easily solved. Based on analysis of past 10 years:
- Case 1 (Distinct Linear): 35% of problems
- Case 2 (Repeated Linear): 25% of problems
- Case 3 (Irreducible Quadratics): 30% of problems
- Case 4 (Mixed Types): 10% of problems
The Partial Fractions Method: Step-by-Step
Step 1: Check if Proper Fraction
If degree of numerator ≥ degree of denominator, perform polynomial division first.
Step 2: Factor Denominator Completely
Factor into linear and irreducible quadratic factors.
Step 3: Set Up Partial Fractions
Based on the factors in denominator:
- Distinct linear factors: $\frac{A}{ax+b}$
- Repeated linear factors: $\frac{A}{ax+b} + \frac{B}{(ax+b)^2} + \cdots$
- Irreducible quadratics: $\frac{Ax+B}{ax^2+bx+c}$
Step 4: Solve for Constants
Use substitution method or equate coefficients.
Step 5: Integrate Each Term
Integrate each partial fraction separately.
Distinct Linear Factors
Evaluate $\int \frac{x+1}{(x-1)(x+2)} dx$
Solution:
Step 1: Set up partial fractions:
$\frac{x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$
Step 2: Multiply through by denominator:
$x+1 = A(x+2) + B(x-1)$
Step 3: Solve for A and B:
Let $x=1$: $2 = A(3) \Rightarrow A = \frac{2}{3}$
Let $x=-2$: $-1 = B(-3) \Rightarrow B = \frac{1}{3}$
Step 4: Rewrite integral:
$\int \left(\frac{2/3}{x-1} + \frac{1/3}{x+2}\right) dx$
Step 5: Integrate:
$\frac{2}{3}\ln|x-1| + \frac{1}{3}\ln|x+2| + C$
Repeated Linear Factors
Evaluate $\int \frac{x^2+1}{x(x-1)^2} dx$
Solution:
Step 1: Set up partial fractions:
$\frac{x^2+1}{x(x-1)^2} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$
Step 2: Multiply through:
$x^2+1 = A(x-1)^2 + Bx(x-1) + Cx$
Step 3: Solve for constants:
Let $x=0$: $1 = A(1) \Rightarrow A = 1$
Let $x=1$: $2 = C(1) \Rightarrow C = 2$
Coefficient of $x^2$: $1 = A + B \Rightarrow B = 0$
Step 4: Rewrite and integrate:
$\int \left(\frac{1}{x} + \frac{2}{(x-1)^2}\right) dx = \ln|x| - \frac{2}{x-1} + C$
Irreducible Quadratic Factors
Evaluate $\int \frac{2x^2+3x+1}{(x+1)(x^2+1)} dx$
Solution:
Step 1: Set up partial fractions:
$\frac{2x^2+3x+1}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+1}$
Step 2: Multiply through:
$2x^2+3x+1 = A(x^2+1) + (Bx+C)(x+1)$
Step 3: Solve for constants:
Let $x=-1$: $0 = A(2) \Rightarrow A = 0$
Coefficient of $x^2$: $2 = A + B = B \Rightarrow B = 2$
Coefficient of $x$: $3 = B + C = 2 + C \Rightarrow C = 1$
Step 4: Rewrite and integrate:
$\int \frac{2x+1}{x^2+1} dx = \int \frac{2x}{x^2+1} dx + \int \frac{1}{x^2+1} dx$
$= \ln|x^2+1| + \tan^{-1}x + C$
🚀 Quick Solving Strategies
For Proper Fractions:
- Always factor denominator completely first
- Use cover-up method for distinct linear factors
- Set up correct form based on factor types
- Check your decomposition by recombining
For Improper Fractions:
- Perform polynomial division first
- Work with remainder as proper fraction
- Don't forget to integrate the polynomial part
- Watch for cancellation opportunities
More Cases & Advanced Problems Available
Includes repeated quadratic factors, improper fractions, and JEE Advanced level problems with detailed solutions
📝 Quick Self-Test
Try these JEE-level integration problems:
1. $\int \frac{3x+2}{x^2-4} dx$
2. $\int \frac{x^3}{x^2-1} dx$ (Improper fraction)
3. $\int \frac{1}{x(x^2+1)^2} dx$
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