Partial Fractions
Partial fraction decomposition splits a complicated rational expression into simple fractions that are easy to integrate or invert.
The method
- Factor the denominator.
- Write one fraction for each factor, with an unknown numerator.
- Multiply through by the denominator and solve for the unknowns.
Worked example
Decompose (3x + 5) / ((x + 1)(x + 2)). Write A/(x + 1) + B/(x + 2), so 3x + 5 = A(x + 2) + B(x + 1).
- Set x = −1: 2 = A(1), so A = 2.
- Set x = −2: −1 = B(−1), so B = 1.
The result is 2/(x + 1) + 1/(x + 2). Check: (2(x + 2) + (x + 1)) / ((x + 1)(x + 2)) = (3x + 5) / ((x + 1)(x + 2)). ✓
Special cases
A repeated factor (x + 1)² needs terms A/(x + 1) + B/(x + 1)². An irreducible quadratic needs a linear numerator Bx + C. If the numerator's degree is not lower than the denominator's, divide first.
Common mistake
Skipping the degree check: decomposition only works on proper fractions.
Why it is useful
Decomposition turns hard integrals into easy ones. For 1/(x(x + 1)) write A/x + B/(x + 1), so 1 = A(x + 1) + Bx. Setting x = 0 gives A = 1, and x = −1 gives B = −1. The integral is then ln|x| − ln|x + 1| + C. The same technique inverts Laplace transforms and sums telescoping series.