Partial Fraction Decomposition
Created | Updated Sep 28, 2008
Partial fraction decomposition is a handy thing to know for calculus, especially when dealing with integrals, as it tends to make them much simpler. What, you may ask, is a partial fraction, and how does one decompose it?
Essentially, the fractions we'll be dealing with here are those of the form f(x)/g(x), where f and g are polynomial functions.
How does it work?
We shall call the degree of the numerator "N", and the degree of the denominator "D." Partial fraction decomposition will only work when N is less than D, so if this is not the case you will have to use long division first to change your fraction from the form f(x)/g(x) to the form h(x)+p(x)/q(x), and continue with your p(x)/q(x) fraction. We'll use the fraction (9x+1)/(x3+x). So, our steps are:
1. Factor the denominator. Our example becomes (9x+1)/x(x2+1)
2. Write that the original fraction equals the sum of a series of other fractions with variables as the numerators and each with one factor of the denominator as their denominators. Some important notes about this: The degree of the numerator of each new fraction should be one less than the degree of its denominator, and any factors raised to a power have to be written multiple times in increasing degrees, starting from one and stopping at the degree they are in the original fraction. Our example is now (9x+1)/x(x2+1)=A/x+(Bx+C)/(x2+1)
3. Multiply by the denominator of the original fraction. Our example becomes 9x+1=A(x2+1)+Bx2+Cx
4. Rearrange so each power of x is with its coefficient:
9x+1=x2(A+B)+Cx+A
5. Now, the coefficients on the left equal the corresponding coefficients on the right, so:
A+B=0
C=9
A=1
6. Solve and rewrite. A=1, B=-1, C=9, so (9x+1)/(x3+x)=1/x+(-x+9)/(x2+1)
