Partial Fraction Decomposition

Partial fraction decomposition rewrites a rational expression as a sum of simpler fractions. You use it after factoring the denominator, usually to make integration or algebra work easier.

The first check is important: the numerator degree must be smaller than the denominator degree. When that condition holds, the expression is called proper. If it does not hold, do polynomial division first and then decompose the remainder.

What Partial Fraction Decomposition Means

A rational expression is a quotient of polynomials, such as

$$\frac{5x + 7}{(x + 1)(x + 2)}.$$

Partial fraction decomposition asks whether that single fraction can be rewritten as a sum like

$$\frac{A}{x + 1} + \frac{B}{x + 2}.$$

If the two forms are equal for all allowed $x$, then the constants $A$ and $B$ capture the same expression in a simpler structure.

When You Can Use Partial Fraction Decomposition

This method works for rational expressions after the denominator is factored over the number system you are using. In most early calculus courses, that means factoring over the real numbers.

The denominator shape tells you the fraction shape:

$$(x-a)(x-b) \quad \Rightarrow \quad \frac{A}{x-a} + \frac{B}{x-b}$$

That structure is the whole idea. If the factorization is wrong or incomplete, the setup will also be wrong.

Worked Example: Decompose A Rational Expression

Decompose

$$\frac{5x + 7}{(x + 1)(x + 2)}.$$

Because the denominator has two distinct linear factors, start with

$$\frac{5x + 7}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}.$$

Multiply both sides by $(x + 1)(x + 2)$ to clear denominators:

$$5x + 7 = A(x + 2) + B(x + 1).$$

Expand the right-hand side:

$$5x + 7 = (A + B)x + (2A + B).$$

Now match coefficients on both sides:

$$A + B = 5$$ $$2A + B = 7$$

Subtract the first equation from the second:

$$A = 2.$$

Then

$$B = 3.$$

So the decomposition is

$$\frac{5x + 7}{(x + 1)(x + 2)} = \frac{2}{x + 1} + \frac{3}{x + 2}.$$

You can verify it by combining the right-hand side again:

$$\frac{2}{x + 1} + \frac{3}{x + 2} = \frac{2(x + 2) + 3(x + 1)}{(x + 1)(x + 2)} = \frac{5x + 7}{(x + 1)(x + 2)}.$$

How The Setup Changes With Different Denominators

The setup always comes from the factors in the denominator.

If the denominator has distinct linear factors, use constant numerators:

$$\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}.$$

If a linear factor repeats, include every power up to that repeat:

$$\frac{P(x)}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}.$$

If a quadratic factor cannot be factored further over the real numbers, use a linear numerator:

$$\frac{P(x)}{x^2 + 1} = \frac{Ax + B}{x^2 + 1}.$$

That last case causes a lot of mistakes. A constant numerator is not enough in general for an irreducible quadratic factor.

Common Partial Fraction Mistakes

1. Skipping the degree check. If the fraction is improper, partial fractions should come after polynomial division, not before.
2. Forgetting repeated factors. For $(x-1)^3$, you need terms for $(x-1)$, $(x-1)^2$, and $(x-1)^3$.
3. Using only constants over an irreducible quadratic factor. Over the real numbers, the numerator should be linear.
4. Solving for constants but not checking the result by recombining the fractions.

Where Partial Fraction Decomposition Is Used

This method appears most often in calculus and algebra. In calculus, it is especially useful for integrating rational functions after the denominator has been factored. In algebra, it can make a rational expression easier to simplify or compare.

The exact form depends on what counts as a factor in your course. For example, a quadratic that stays irreducible over the real numbers may factor over the complex numbers, and that would change the decomposition.

It All Comes From the Denominator

Every part of partial fraction decomposition follows from how the denominator factors: distinct linear factors take constant numerators, a repeated factor needs a term for each power, and an irreducible quadratic needs a linear numerator. Factor correctly, match the form to those factors, solve for the constants, and recombine to check. That sequence is the whole method, and it is what makes integrating rational functions manageable in calculus.