SOLUTION: Resolve the following improper fraction into partial fractions: {{{x^4/(x^2+1)(x^2-1)}}}

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Question 276535: Resolve the following improper fraction into partial fractions:


x%5E4%2F%28x%5E2%2B1%29%28x%5E2-1%29
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
The phrase "partial fractions" suggests that you are either in a Calculus class or in a Math class which is preparing you with some of the Algebra skills needed in Calculus. If I am wrong about all this then what follows may be either way too much or way off base.

What needs to be done is:
  1. Multiply out the denominator
  2. Divide out the fraction giving a quotient and a (proper) remainder fraction
  3. Build a sum of partial fractions which add up to the remainder fraction.

1. Multiply out the denominator:
x%5E4%2F%28x%5E4-1%29

2. Divide.
         1
         _______
x^4 - 1 /x^4
         x^4 - 1
         -------
               1

So x%5E4%2F%28x%5E4-1%29+=+1+%2B+1%2F%28x%5E4-1%29

3. Build the partial fractions for 1%2F%28x%5E4-1%29. For this we need to factor the denominator:
1%2F%28x%5E2%2B1%29%28x-1%29%28x%2B1%29
There is potentially a fraction for each factor of the denominator. And the numerator for each denominator will be a polynomial of degree which is one less than the degree of its denominator. Here's the prototype:

If we were to add the fractions on the right, we should get the fraction on the left. Adding the fractions on the right requires that we get the denominators the same. This requires that the numerator and denominator of each fraction be multiplied by the product of the other two denominators:


For now I am going to focus on just the numerators. (We know the denominators will all work out to be x%5E4-1). Multiplying out the numerators we get:

The big mess needs to add up to 1%2F%28x%5E4-1%29. So in the expression above, the coefficients of the x%5E3 terms, the x%5E2 terms and the x terms must all by zero since there are no x's in the desired numerator of 1. And the constant (x-less) terms must add up to 1:
x%5E3 coefficients: A + C + D = 0
x%5E2 coefficients: B + C - D = 0
x coefficients: -A + C + D = 0
Constants: -B + C - D = 1

The above equations are a system of 4 equations with 4 variables. There are a variety of ways to solve such a system (Substitution, Linear Combination (aka Elimination), and various matrix/determinant-based methods. The solution works out to be:
A = 0
B = -1/2
C = 1/4
D = 1/2
Substituting these into

we get:

which simplifies to:

Each term in the final expression on the right is in an integrable form.