Partial Fraction Setup
Decompose a rational expression into simpler fractions.
Formula
Variables
Example
Used before integrating rational functions
How it works
When a rational expression has a denominator that factors into two distinct linear terms, it can often be rewritten as a sum of simpler fractions with those same denominators. This works because after multiplying both sides by (x-a)(x-b), the identity must hold for every x, so the numerators can be chosen to match exactly. The constants A and B are found by equating coefficients or plugging in convenient values of x.
When to use it
Use this setup whenever you need to break apart a rational function with a factored denominator, especially in integration, inverse Laplace transforms, or solving algebraic equations involving rational expressions. It is most effective when the denominator splits into distinct linear factors.
Worked example, step by step
Decompose 1 / ((x-2)(x+3)) Step 1: Write the setup: 1 / ((x-2)(x+3)) = A / (x-2) + B / (x+3) Step 2: Multiply through by (x-2)(x+3): 1 = A(x+3) + B(x-2) Step 3: Substitute x = 2: 1 = A(5) + B(0), so A = 1/5 Step 4: Substitute x = -3: 1 = A(0) + B(-5), so B = -1/5 Step 5: Final decomposition: 1 / ((x-2)(x+3)) = (1/5)/(x-2) - (1/5)/(x+3)
Common mistakes to avoid
- Forgetting to multiply both sides by the full denominator before solving for A and B
- Using the wrong sign when substituting a or b into the linear factors
- Assuming the decomposition works without checking that the denominator factors into distinct linear terms
Did You Know?
Partial fractions break a messy rational expression into simple pieces — indispensable for integration and control theory.
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