Partial Fraction Decomposition

Partial fraction decomposition is a technique used to rewrite a rational function as a sum of simpler rational expressions, each with a simpler denominator.

This method is especially powerful when it comes to evaluating integrals involving rational functions.

A Practical Example

Let’s examine the rational function:

$$ \frac{1}{x^3 - 3x^2 + 2x} $$

We start by factoring out $ x $ from the denominator:

$$ \frac{1}{x \cdot (x^2 - 3x + 2)} $$

The roots of the denominator are $ x = 0 $, $ x = 1 $, and $ x = 2 $.

To factor the quadratic $ x^2 - 3x + 2 $, we apply Ruffini’s synthetic division:

$$ \begin{array}{c|lcc|r} & 1 & -3 & 2 & 0 \\ 1 & & 1 & -2 & 0 \\ \hline & 1 & -2 & 0 & 0 \end{array} $$

This confirms the factorization:

$$ x^2 - 3x + 2 = (x - 1)(x - 2) $$

So the original expression becomes:

$$ \frac{1}{x(x - 1)(x - 2)} $$

We now decompose it into partial fractions:

$$ \frac{1}{x(x - 1)(x - 2)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x - 2} $$

where $ A $, $ B $, and $ C $ are constants to be determined.

Multiplying both sides by the common denominator yields:

$$ 1 = A(x - 1)(x - 2) + Bx(x - 2) + Cx(x - 1) $$

We now expand the right-hand side:

$$ A(x^2 - 3x + 2) + B(x^2 - 2x) + C(x^2 - x) $$

Combining like terms gives:

$$ (A + B + C)x^2 + (-3A - 2B - C)x + 2A $$

Matching coefficients on both sides (since the left-hand side is 1), we obtain the following system:

$$ \begin{cases} A + B + C = 0 \\ -3A - 2B - C = 0 \\ 2A = 1 \end{cases} $$

Explanation. The coefficient of $ x^2 $ on the left is 0, so $ A + B + C = 0 $. Similarly, the coefficient of $ x $ is 0, yielding $ -3A - 2B - C = 0 $. Finally, the constant term is 1, which gives $ 2A = 1 $.

Solving the system by substitution:

From the third equation: $ A = \frac{1}{2} $

Substitute into the first equation:

$$ \frac{1}{2} + B + C = 0 \Rightarrow B + C = -\frac{1}{2} $$

Substitute into the second equation:

$$ -3 \cdot \frac{1}{2} - 2B - C = 0 \Rightarrow -\frac{3}{2} - 2B - C = 0 $$

Now solve the system:

From the first equation: $ C = -\frac{1}{2} - B $

Plug into the second:

$$ -\frac{3}{2} - 2B - (-\frac{1}{2} - B) = 0 $$

Which simplifies to:

$$ -\frac{3}{2} + \frac{1}{2} - B = 0 \Rightarrow -1 - B = 0 \Rightarrow B = -1 $$

Then:

$$ C = -\frac{1}{2} - (-1) = \frac{1}{2} $$

So the constants are:

$$ A = \frac{1}{2}, \quad B = -1, \quad C = \frac{1}{2} $$

Substituting back, the decomposition is:

$$ \frac{1}{x(x - 1)(x - 2)} = \frac{1}{2x} - \frac{1}{x - 1} + \frac{1}{2(x - 2)} $$

This expresses the original rational function as a sum of simple fractions:

$$ \frac{1}{2x} - \frac{1}{x - 1} + \frac{1}{2(x - 2)} $$

Why is this useful? Partial fraction decomposition is a powerful tool for integrating rational functions. Now that the integrand has been rewritten, we can evaluate the integral more easily: $$ \int \left( \frac{1}{2x} - \frac{1}{x - 1} + \frac{1}{2(x - 2)} \right) dx $$ This becomes a sum of basic integrals: $$ \frac{1}{2} \int \frac{1}{x} \, dx - \int \frac{1}{x - 1} \, dx + \frac{1}{2} \int \frac{1}{x - 2} \, dx $$ Evaluating each term: $$ \frac{1}{2} \log |x| - \log |x - 1| + \frac{1}{2} \log |x - 2| + c $$

And so on.