Partial Fraction Expansion

When working with fractions, we often find common denominators in order to be able to add and subtract fractions, to help simplify equations. Partial fractions is the reverse process of this. Often, we will encounter integrals that are rational functions we don’t know how to take the integral of. Using partial fractions, we can decompose the single rational function into a sum of rational functions that are easy to integrate, such that both functions are equivalent.

Example: Solve \int \frac{x+5}{x^2+x-2} dx

Currently, we don’t have a good way to solve this integral, however, we can simplify the rational function to get: \frac{x+5}{x^2+x-2} = \frac{2}{x-1} - \frac{1}{x+2}.

From here, we can easily solve the integral, using the fact that \int \frac{1}{f(x)} = ln|f(x)|

\int \frac{2}{x-1} - \frac{1}{x+2} = 2*\int \frac{1}{x-1} - \int \frac{1}{x+2} = 2*ln|x-1| - ln|x+2| + C

There are a number of ways we can decompose rational functions into partial fractions, depending on the rational function we are working with. Suppose that we have a rational function \frac{P(x)}{Q(x)}. The following cases for partial fractions exist.

Case 1: The denominator, Q(x), is a product of distinct linear factors.

This means that we can rewrite Q(x) as Q(x) = (a_1x+b_1)(a_2x+b_2) \dots (a_kx+b_k), where no factor is repeated.

In this situation, there exists constant values A_1,A_2, \dots, A_k such that:

\frac{R(x)}{Q(x)} = \frac{A_1}{a_1x+b_1} + \frac{A_2}{a_2x+b_2} + \dots + \frac{A_k}{a_kx_b_k}

Example: Evaluate \int \frac{x^2+2x-1}{2x^3+3x^2-2x} dx

First, we need to rewrite Q(x) as a product of linear factors: 2x^3+3x^2-2x = x(2x^2+3x-2) = x(2x-1)(x+2)

Now that we know the linear terms, we can determine the constants for the top portions of the fractions.

\frac{x^2+2x-1}{2x^3+3x^2-2x} = \frac{A}{x} + \frac{B}{2x-1} + \frac{C}{x+2}

To determine the values of A, B, and C, we just need to determine which values result in the function x^2 + 2x -1 when we combine the three partial fractions.

More specifically, we need x^2+2x-1 = A(2x-1)(x+2) + B(x)(x+2) + C(x)(2x-1). We can simplify this equation to get a system of equations.

A(2x-1)(x+2) + B(x)(x+2) + C(x)(2x-1)
=2Ax^2+Bx^2+2Cx^2 + 3Ax+2Bx-Cx-2A
=(2A+B+2C)x^2 + (3A+2B)x - 2(A)

From here, we just need to solve the three equations:

x^2 = (2A+B+2C)x^2
2x = (3A+2B)x
-1 = -2A

From the third equation, we see that A = \frac{1}{2}. We then substitute this into the second equation to get 2x = (\frac{3}{2}+2B)x. Solving this gives us B = \frac{1}{5}. Finally, the first equation will give us C = \frac{-1}{10}

This means that our partial fraction expansion is: \int \frac{1}{2} \frac{1}{x} + \frac{1}{5}\frac{1}{2x-1} - \frac{1}{10} \frac{1}{x+2} dx = \frac{1}{2} ln|x| + \frac{1}{10} ln|2x-1| - \frac{1}{10} ln|x+2| + C

Case 2: The denominator Q(x) is a product of linear factors, some of which are repeated.

When we reduce to repeated terms, we just need to have a fraction for each term that is repeated, raised to the power of its repetition.

For instance, suppose we had (x+1)(x-1)(x-1). We repeat the term (x-1) twice, which means that we will have the fraction terms \frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}

From here, we just solve for A, B, and C, in the exact same way as case 1.

Case 3: Q(x) contains irreducible quadratic factors, none of which are repeated.

Quadratic irreducible terms in our partial fractions will alter the numerator of the fraction rather than the denominator. To compensate for quadratic terms, the typical constant A becomes a linear constant, Ax+B.

For instance, suppose we had the terms (x-2), (x^2+1), (x^2+4). In this case, we would have a partial fraction of \frac{A}{x-2} + \frac{Bx+C}{x^2+1}+\frac{Dx+E}{x^2+4}.

Once again from here, you just solve for the unknowns to get the partial fraction expansion.

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