Integration of Rational Functions by Partial Fractions

Some rational functions are too complicated to integrate directly, but they can be rewritten as a sum of simpler fractions that each integrate easily. This technique, called integration by partial fractions, relies on partial fraction decomposition from algebra: breaking a fraction like \(\frac{2x+3}{(x-1)(x+2)}\) into pieces like \(\frac{A}{x-1}+\frac{B}{x+2}\). Once the fraction is split apart, each term becomes a basic integral you already know how to solve.

Partial fractions work on integrals of the form \(\int \frac{P(x)}{Q(x)}\,dx\), where \(P(x)\) and \(Q(x)\) are polynomials. Before decomposing, check the degrees:

• If the degree of \(P(x)\) is less than the degree of \(Q(x)\), the fraction is "proper" and ready to decompose.
• If the degree of \(P(x)\) is greater than or equal to the degree of \(Q(x)\), divide first using polynomial long division, then decompose the leftover proper fraction.

This method pairs with other integration strategies you may already know, such as u-substitution, which is often used once the fraction has been split into simpler pieces.

How you set up the decomposition depends on how the denominator factors.

Case 1: Distinct Linear Factors

If \(Q(x)\) factors into distinct linear pieces, such as \(Q(x) = (x-a)(x-b)\), write:

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

Multiply both sides by the denominator and solve for \(A\) and \(B\) by substituting convenient values of \(x\) or by comparing coefficients. Sometimes the factoring step itself uses the difference of squares formula when the denominator looks like \(x^2 - a^2\).

Case 2: Repeated Linear Factors

If a linear factor repeats, such as \((x-a)^2\), include one term for every power up to that repetition:

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

Case 3: Irreducible Quadratic Factors

If \(Q(x)\) contains a quadratic factor \(x^2+bx+c\) that cannot be factored into real linear pieces, its numerator must be linear rather than constant:

\(\frac{P(x)}{x^2+bx+c} = \frac{Ax+B}{x^2+bx+c}\)

Terms like this often integrate to a combination of a logarithm and an arctangent function.

1. Confirm the fraction is proper (divide first if it is not).
2. Factor the denominator completely into linear and irreducible quadratic pieces.
3. Write the correct partial fraction form for each factor type.
4. Clear denominators and solve for the unknown constants.
5. Rewrite the original integral as a sum of simpler integrals.
6. Integrate each term using logarithm, power, or arctangent rules.

Evaluate \(\int \frac{3x+5}{(x-1)(x+2)}\,dx\).

Set up the decomposition: \(\frac{3x+5}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2}\).

Clearing denominators gives \(3x+5 = A(x+2) + B(x-1)\). Letting \(x=1\) gives \(8 = 3A\), so \(A = \frac{8}{3}\). Letting \(x=-2\) gives \(-1 = -3B\), so \(B = \frac{1}{3}\).

The integral becomes:

\(\int \frac{8/3}{x-1}\,dx + \int \frac{1/3}{x+2}\,dx = \frac{8}{3}\ln|x-1| + \frac{1}{3}\ln|x+2| + C\)

• Forgetting to divide first when the numerator's degree is not smaller than the denominator's degree.
• Using a constant numerator over an irreducible quadratic factor instead of a linear one.
• Missing a required term when a linear factor is repeated.
• Dropping the absolute value inside the logarithm after integrating a linear factor.

Partial fractions is one of several tools for handling integrals that resist basic formulas; others include improper integrals for unbounded limits or discontinuities, and trigonometric substitution for expressions involving square roots of quadratics. Choosing the right technique starts with recognizing the structure of the integrand.

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