When adding two fractions a/b + c/d one first computes the common denominator, b.d, and then combines fractions over the common denominator a/b + c/d = (ad+bc)/(bd) Partial fractions is essentially the opposite procedure, one reduces a more complicated quotient into the sum of simpler quotients.

As a simple example, take the expression

1 / (x²-1) After factoring the denominator we can write this as 1/((x+1)(x-1)) Expressing this as the sum of two simpler fractions 1/((x+1)(x-1)) = A/(x+1) + B/(x-1) where A and B are constants, the only thing that remains is to solve for A and B. We can do this by recombining the right hand side A/(x+1) + B/(x-1) = (A(x-1)+B(x+1))/((x+1)(x-1)) and equating numerators A(x-1) + B(x+1) = 1 This gives (A+B)x + (B-A) = 1 or A+B=0
B-A=1 The second equation gives B=A+1, which when substituted into the first equation gives 2A+1=0 => A=-1/2
B=1/2 or Note, another, somewhat quicker way, is to note that the equation A(x-1) + B(x+1) = 1 is valid for all x , and consequently, if x=1, the first term disappears, leaving B*2=1, or B=1/2. If x=-1, the second term disappears, leaving A*(-2)=1, or A=-1/2, which is the same result as before.

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There are four distinct cases, in which the denominator is composed of:

1. distinct linear factors
2. repeated linear factors
3. distinct non-factorable (irreducible) quadratics
4. repeated non-factorable (irreducible) quadratics

In the first case, we have In the second case, where the factor (a₁x+b₁) is appears to the k^th power, the terms are generated. In the third case, we have and finally in the fourth case, where the quadratic term appears to the k^th power, the terms appear.

If the degree of the numerator R(x) is greater than or equal to the denominator Q(x), then synthetic division must be used.

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Note: the linear terms will integrate to natural logs, the quadratic terms will integrate to natural logs or inverse trig functions, and higher terms will be somewhat more complicated.

Note: On an exam, a standard problem is to find the form of the partial fraction decomposition, but not to find the actual coefficients.

Copyright © 1997, Michael S. Pilant