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Partial Fraction Decomposition
Calculus 2 · Chapter 1: Review & Advanced Integration TechniquesSection 1.5 · lesson 5 of 5413 min · free
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In this lesson
Splitting one hard fraction into several easy ones.
- Test properness, and divide first when the fraction is improper
- Match the decomposition form to the denominator's factor types
- Solve for constants by strategic substitution and coefficient matching
- Integrate every piece as a logarithm, a power, or an arctangent
Transcript
Everything said in the video, section by section (1,834 words). Read along, or search it.
MAT-265 - Calculus 2 · Section 1.5
Partial Fraction Decomposition
A biologist is modeling how a trait spreads through a population. They're using the logistic growth model — d P d t equals k P times one minus P. To predict the population at any future time, that equation has to be integrated. Separate the variables, and you get the integral of d P over P times one minus P. And that denominator — a product of two factors — resists everything you've learned. No substitution simplifies it. No trigonometric identity applies. So here's the breakthrough. Instead of attacking the fraction as one object, split it into a sum of simpler fractions. Each of which integrates to a logarithm. That splitting process is called partial fraction decomposition. And it unlocks the logistic model, electrical circuit analysis, and a great deal more.
The Problem
A Flow That Fights Itself
Let's be precise about what this technique actually is. A rational function is just a ratio of two polynomials. P of x over Q of x. And here's the nice way to think about decomposition. You already know how to add fractions over a common denominator. You've done it since middle school. Partial fraction decomposition runs that process backward. It takes a single rational function and rewrites it as a sum of fractions whose denominators are the factors of Q. Because each resulting piece is simple enough to integrate directly, one hard integral becomes several easy ones. That's the whole idea. Everything else is procedure.
Step Zero
Proper or Improper? Check Degrees First
Before you decompose anything, there's a test you have to run. Decomposition applies only to a proper rational function. Proper means the degree of the numerator is strictly less than the degree of the denominator. If the numerator's degree is greater than or equal to the denominator's, the fraction is improper — and you must do polynomial long division first. Long division rewrites an improper fraction as a polynomial, plus a proper remainder fraction. The polynomial part integrates trivially. The remainder is what gets decomposed. Let's see it. Prepare x cubed plus x, over x squared minus one. Numerator degree three, denominator degree two. Improper. So divide. x cubed plus x equals x squared minus one, times x, plus two x. So the whole thing becomes x, plus two x over x squared minus one. And that remainder — degree one over degree two — is proper, and ready to decompose. So here's your first move, every single time. Compare degrees. Numerator smaller? Go straight to decomposition. Otherwise, divide first.
Case 1
Distinct Linear Factors
Once the fraction is proper, factor the denominator completely. The form of the decomposition is dictated entirely by what those factors look like. Three cases cover every possibility. Case one. Distinct linear factors. If the denominator factors into distinct, non-repeated linear terms, assign a single constant numerator to each one. Let's do the logistic integral. One over x times one minus x. The denominator is already factored into two distinct linear pieces. So set it equal to A over x, plus B over one minus x. Multiply both sides by x times one minus x to clear denominators. You get one equals A times one minus x, plus B x. Now here's the efficient move. Strategic substitution. Pick x-values that zero out a term. Let x be zero: one equals A, so A is one. Let x be one: one equals B, so B is one. Two constants, two lines, no algebra system. Integrate each piece and you get log absolute x, minus log absolute one minus x. Now — where did that minus sign come from? Not from the decomposition. From the chain rule. The derivative of one minus x is negative one. That's the most common slip in Case one. Watch for it.
Case 2
Repeated Linear Factors
Case two. Repeated linear factors. Suppose a linear factor appears to a power — a x plus b, all to the k. One fraction is not enough here. You need a separate term for every power, from one all the way up to k. And this matters: omitting any intermediate power leaves the system unsolvable. Include them all, even if you suspect a coefficient is zero. Try the integral of three x plus five, over x plus one, squared. The denominator is a linear factor repeated twice, so use two terms. A over x plus one, plus B over x plus one squared. Multiply through. You get three x plus five equals A times x plus one, plus B. Let x equal minus one to kill A. That gives B equals two. For A, match coefficients of x. The left side has three x, the right has A x. So A is three. Now integrate — and notice you get two different kinds of term. The first gives a logarithm. The second is a power-rule integral, giving minus two over x plus one. Repeated factors always mix logarithms with power terms. Expect both.
Case 3
Irreducible Quadratic Factors
Case three. Irreducible quadratic factors. A quadratic is irreducible over the real numbers when it can't be factored into real linear pieces. Equivalently — and this is the test you'll actually use — when its discriminant, b squared minus four a c, is negative. Each irreducible quadratic factor gets a linear numerator. A x plus B. Not just a constant. Why a linear numerator? Because the denominator is degree two, so the numerator can carry degree one and still be proper. Let's do one. The integral of two x squared minus x plus four, over x times x squared plus four. The denominator has a linear factor x, and the quadratic x squared plus four — discriminant zero minus sixteen, so irreducible. Set up a constant over the linear factor, and a linear numerator over the quadratic. Clear denominators, substitute x equals zero, and A falls out immediately as one. Then expand and match coefficients. The x squared terms give A plus B equals two, so B is one. The x terms give C equals minus one. Now here's the question worth sitting with. Substituting x equals zero instantly gave you A. But no real value of x makes x squared plus four vanish. That is precisely what irreducible means. So strategic substitution alone can never determine B and C. That's when equating coefficients stops being convenient and becomes necessary.
Reference
Match the Term to the Factor
Let's collect all three cases into one reference, because this is how you'll use it in practice. Look at each factor in the denominator, and read off what it contributes. A distinct linear factor contributes one constant over it, and integrates to a logarithm. A repeated linear factor contributes a chain of terms, one per power, and integrates to a logarithm plus power-rule terms. An irreducible quadratic contributes a linear numerator over it, and integrates to a logarithm plus an arctangent. And a repeated irreducible quadratic escalates the same way, giving logs, arctangents and power terms. For finding the constants, you have two complementary methods. Strategic substitution. Plug in the roots of the denominator. Each root zeroes all but one term, exposing one constant immediately. Fastest for linear factors. Equating coefficients. Expand the right side fully and match each power of x, producing a linear system. That's the reliable workhorse when substitution can't reach every constant — especially with irreducible quadratics, whose roots aren't real. In practice you combine them. Substitution grabs the easy constants, then coefficient matching mops up the rest.
Integrating the Pieces
Logarithms, Powers, and Arctangents
The payoff of all this decomposition is that every resulting term matches one of a small set of standard integrals. Know these by sight and the final step becomes mechanical. A linear factor gives a logarithm. A over a x plus b integrates to A over a, times log absolute a x plus b. A repeated linear factor gives the power rule. A over a x plus b, to the n, with n at least two. And a quadratic factor gives a logarithm plus an arctangent. The A x part integrates to a logarithm, via the substitution u equals x squared plus a squared. The constant part integrates to an arctangent, using the standard form: the integral of d x over x squared plus a squared is one over a, arctangent of x over a. But there's a catch. That arctangent form needs the denominator in the clean shape x squared plus a squared. When it isn't, complete the square. Take the integral of d x over x squared plus two x plus five. Discriminant four minus twenty — irreducible. Complete the square: x plus one, squared, plus four. Let u be x plus one, and a is two. The answer is one half arctangent of x plus one over two, plus C.
The Universal Recipe
A Linear Numerator Over a Completed Square
Now let's combine both quadratic skills into one problem, because this is the pattern you'll meet most often. Evaluate the integral of three x plus one, over x squared plus four x plus thirteen. A linear numerator, and a denominator that has to be completed. First check the discriminant: sixteen minus fifty-two. Negative, so it's irreducible and no factoring is possible. Complete the square in the denominator: x plus two, squared, plus nine. Now here's the strategy for a linear numerator, and it's worth learning as a shape rather than a formula. You engineer the derivative of the denominator into the top. The derivative of x squared plus four x plus thirteen is two x plus four. So rewrite the numerator, three x plus one, as a multiple of two x plus four, plus a constant correction. Three halves times two x plus four gives three x plus six. And three x plus six, minus five, is three x plus one. So the correction is minus five. Split the integral accordingly. The first piece is a logarithm, by substitution. The second is an arctangent, with a equals three. Derivative-of-the-denominator piece, plus leftover-constant piece. That split is the universal recipe for any linear numerator over an irreducible quadratic. Learn the shape.
Key Takeaways
Why This Always Works
Before the takeaways, it's worth pausing on why this method always works. There's a foundational result in algebra: any polynomial with real coefficients factors completely into real linear and irreducible quadratic factors. The three cases you've just learned cover exactly those two factor types, including their repeated versions. So there is no proper rational function this method cannot decompose. It's not a collection of tricks. It's a complete procedure. Factor, set up the matching form, solve for constants, integrate piece by piece. So — five things to carry forward. Check degrees first. Decompose only proper fractions; otherwise do long division before anything else. Factor the denominator completely — the factor types determine the form. Match the term to the factor. A constant over each linear factor, an escalating chain for repeated factors, a linear numerator over each irreducible quadratic. Solve coefficients by combining methods. Substitution for the roots, coefficient matching for the rest. And integrate by pattern. Linear factors give logarithms, repeated factors give power-rule terms, irreducible quadratics give a logarithm plus an arctangent. Next up is section one point six, Integration Strategy — which ties every technique in this chapter into one decision procedure.
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