What are Calculus II · Integration

Partial Fractions?

A powerful integration technique that breaks complex rational functions into simpler pieces — making them easy to integrate using basic logarithm and arctangent rules.

Partial fraction decomposition is a technique used in calculus to integrate rational functions — fractions where the numerator and denominator are polynomials. Instead of trying to integrate a complicated fraction directly, we break it into a sum of simpler fractions that we already know how to integrate.

For example:

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

The right side is much easier to integrate because we know:

  • \(\int \frac{1}{x-a} dx = \ln|x-a| + C\) (for linear factors)
  • \(\int \frac{1}{x^2+a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C\) (for quadratic factors)

Partial fractions turn a difficult integral into a sum of simple integrals.

Integral of
\(\frac{P(x)}{Q(x)}\)
Partial Fractions
\(\frac{A}{x-a} + \frac{B}{x-b} + \cdots\)
Easy Integrals
\(\ln + \tan^{-1}\)
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Breaking apart to rebuild

Think of it like taking apart a complex machine to work on each component separately. The complex fraction is the assembled machine — hard to work with. Partial fractions disassemble it into individual parts (simple fractions), each of which is easy to handle. Then you can integrate each part and put the results together.

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Why it works for integration

When you decompose a rational function, you get terms of the form \(\frac{A}{x-a}\) (which integrate to logarithms) and \(\frac{Ax+B}{x^2+px+q}\) (which integrate to logarithms and arctangents). These are some of the easiest functions to integrate — exactly why partial fractions are so valuable in calculus.

Use partial fractions when integrating a rational function \(\frac{P(x)}{Q(x)}\) where:

The integrand is a rational function (polynomial divided by polynomial)
The degree of the numerator is less than the degree of the denominator (proper fraction)
The denominator can be factored into linear and irreducible quadratic factors
If the fraction is improper (degree of numerator ≥ degree of denominator), first divide using polynomial long division
If the denominator has no real factors, other methods (like trigonometric substitution) may be needed
Distinct linear factors
\(\frac{A}{x-a} + \frac{B}{x-b}\)
Integrates to: \(A\ln|x-a| + B\ln|x-b| + C\)
Repeated linear factors
\(\frac{A}{x-a} + \frac{B}{(x-a)^2}\)
Integrates to: \(A\ln|x-a| - \frac{B}{x-a} + C\)
Irreducible quadratic
\(\frac{Ax+B}{x^2+px+q}\)
Integrates to: \(\ln\) and \(\tan^{-1}\) terms
Repeated quadratic
\(\frac{Ax+B}{x^2+px+q} + \frac{Cx+D}{(x^2+px+q)^2}\)
Requires reduction formulas for higher powers

Follow these steps to integrate a rational function using partial fractions:

Step 1Check if the fraction is proper. If not, divide using polynomial long division.
Step 2Factor the denominator completely into linear and irreducible quadratic factors.
Step 3Set up the partial fraction form based on the types of factors.
Step 4Multiply both sides by the denominator and solve for coefficients (substitute values or equate coefficients).
Step 5Write the integrand as a sum of partial fractions.
Step 6Integrate each term separately:
 •\(\int \frac{A}{x-a} dx = A\ln|x-a| + C\)
 •\(\int \frac{Ax+B}{x^2+px+q} dx =\) combine \(\ln\) and \(\tan^{-1}\)
Example 1 — Linear factors
\(\int \frac{5x-3}{(x-2)(x+1)} dx\)
Decompose: \(\frac{1}{x-2} + \frac{4}{x+1}\)
\(= \ln|x-2| + 4\ln|x+1| + C\)
Example 2 — Repeated linear
\(\int \frac{x}{(x-1)^2} dx\)
Decompose: \(\frac{1}{x-1} + \frac{1}{(x-1)^2}\)
\(= \ln|x-1| - \frac{1}{x-1} + C\)
Example 3 — Quadratic factor
\(\int \frac{1}{x(x^2+1)} dx\)
Decompose: \(\frac{1}{x} - \frac{x}{x^2+1}\)
\(= \ln|x| - \frac{1}{2}\ln(x^2+1) + C\)
Example 4 — Improper fraction
\(\int \frac{x^2+1}{x-1} dx\)
Divide first: \(x+1 + \frac{2}{x-1}\)
\(= \frac{x^2}{2} + x + 2\ln|x-1| + C\)
✗ Forgetting the absolute value in \(\ln\)
\(\int \frac{1}{x-a} dx = \ln|x-a| + C\), not \(\ln(x-a) + C\). The logarithm's argument must be absolute value because the domain is all real numbers except \(a\).
✗ Not checking if the fraction is proper first
If the degree of the numerator is greater than or equal to the denominator, you must perform polynomial long division first. Partial fractions only work for proper rational functions.
✗ Incorrect integration of \(\frac{1}{x^2+a^2}\)
\(\int \frac{1}{x^2+a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C\). Don't forget the \(\frac{1}{a}\) factor! For \(x^2+1\), it's \(\tan^{-1}(x) + C\).
✗ Mixing up \(\ln\) and \(\tan^{-1}\) forms
\(\frac{A}{x-a}\) → \(\ln\) (if denominator is linear)
\(\frac{Ax+B}{x^2+px+q}\) → combination of \(\ln\) and \(\tan^{-1}\) (if denominator is quadratic)

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