Partial Fractions: Clean Integration on the FE Exam

The decomposition rules don’t change.

Linear factors like (ax + b) get A/(ax + b). Repeated factors like (ax + b)² get multiple terms with increasing powers. Irreducible quadratics get (Ax + B) in the numerator, not just a constant.

But recognizing which rule applies when you factor x² + x – 6 into (x + 3)(x – 2)—and setting up A/(x + 3) + B/(x – 2) without second-guessing whether you need more terms—is where partial fractions problems break down under pressure.

The friction starts the moment you finish factoring.

You know you need decomposition terms, but you’re not sure if that factor you just wrote down needs one term or three.

You’re not sure whether to put a constant or a linear expression in the numerator.

You’re not sure which x values to plug in when solving for constants, and choosing wrong means two minutes of algebra that dead-ends.

Partial fractions shows up on the FE when rational expressions refuse to integrate cleanly—Laplace transforms, differential equations, signal processing problems where a complex fraction needs breaking apart before you can extract a meaningful result.

The problem isn’t that the individual algebra steps are hard. It’s that partial fractions has more structural checkpoints than most integration methods—factor completely, assign terms based on factor type, combine fractions correctly, solve for constants strategically—and missing one checkpoint early means the entire setup collapses even if your algebra is perfect.

This guide walks you through a systematic approach that works on any partial fractions problem the FE throws at you—from simple linear factors to repeated quadratics with mixed terms. You’ll learn exactly which decomposition terms to use based on how the denominator factors, how to solve for constants without burning time on messy algebra, and how to verify your setup produces integrals you can actually evaluate.

Before we walk through it step by step, watch this short video. It shows you the full process from checking polynomial degrees to factoring the denominator to assigning the correct decomposition terms to solving for constants to finishing the integral. You’ll see exactly where students typically lose time on partial fractions problems—assigning wrong terms, combining fractions incorrectly, choosing poor x values—and how to avoid those traps completely.

What Is Partial Fractions?

Partial fraction decomposition is the process of breaking a single complex rational expression into a sum or difference of simpler rational expressions that you can integrate individually. Instead of trying to force one complicated integral, you split it into multiple straightforward integrals.

The partial fractions technique only works when the degree of the polynomial in the numerator is strictly less than the degree of the polynomial in the denominator. If that condition isn’t met, you need to perform polynomial long division first to separate the expression into a polynomial plus a proper rational expression.

Think of it like dismantling a machine into its individual parts so you can understand what each piece does. A complex fraction might be impossible to integrate as-is, but once you break it into simpler fractions—each with a linear or simple quadratic denominator—the integration becomes routine.

This matters because many engineering problems produce rational expressions that don’t fit clean substitution patterns. Control systems transfer functions, electrical circuit analysis, and differential equation solutions all generate fractions that require decomposition before you can extract meaningful results.

On the FE Exam, partial fractions shows up when you’re asked to integrate a rational expression where the numerator and denominator don’t have an obvious derivative relationship, or when you need to find inverse Laplace transforms that require breaking a complex fraction into recognizable standard forms.

The method gives you a mechanical process for handling integrals that would otherwise require guessing or pattern recognition under time pressure.

Why Partial Fractions Feel Harder Than They Should

Partial fractions creates friction the moment you finish factoring the denominator and have to decide what decomposition terms to write down.

The uncertainty builds in stages:

• You factor the denominator and get something like (x – 2)(x + 3)², and now you’re supposed to assign terms, but you’re not sure if repeated factors get one term or multiple terms, and whether the exponent matters
• You write down A/(x – 2) + B/(x + 3) but you forgot the repeated factor rule, so you set up the wrong number of terms and don’t realize it until the algebra doesn’t work
• You combine the fractions back together by multiplying through, but you lose track of which terms get which denominators, and the numerators balloon into expressions you can’t simplify cleanly
• You set the numerators equal and stare at an equation with three unknowns and one x, and you’re not sure which x values to plug in to eliminate terms without creating fractions or complex algebra

The real issue isn’t the individual algebra steps. It’s that the partial fractions setup has more structural checkpoints than most integration methods, and each checkpoint has multiple correct-looking options that lead to different setups.

You’re working through a method where the setup phase determines whether the rest of the problem is clean or a mess, and there’s no immediate feedback until you’re already committed to a path.

Breaking Down Partial Fractions Step by Step

Partial fractions problems create uncertainty at the exact moment you factor something like (x – 2)(x + 3)² and have to decide: does that squared term get one decomposition term or two?

Is it just B/(x + 3)² or do I need both B/(x + 3) AND C/(x + 3)²?

That split-second choice—matching factor type to term structure—determines whether your constants will solve cleanly or whether you’ll spend two minutes on algebra that can’t possibly close because you set up the wrong number of unknowns.

Most students either guess at decomposition terms based on what “looks right,” or they freeze trying to remember factor-to-term rules they haven’t used since Calculus II. Both approaches burn time and create setup errors that break the entire problem.

Here’s the workflow that handles any partial fractions problem on the FE. It gives you explicit rules for every factor type—simple linear, repeated linear, simple quadratic, repeated quadratic—so you’re matching factors to terms mechanically, not from memory. Once the terms are set up correctly, solving for constants and integrating becomes straightforward algebra.

This is the same partial fractions process whether you’re decomposing 1/(x² – 1) or something with five factors and mixed quadratics. The decision logic doesn’t change.

Let’s lay out the steps.

Step 1: Check the degree condition

The first thing you need to do is compare the degree of the numerator polynomial to the degree of the denominator polynomial. The degree is the highest exponent that appears in each polynomial.

Partial fraction decomposition only works directly when the numerator degree is strictly less than the denominator degree. If the numerator degree is greater than or equal to the denominator degree, you must perform polynomial long division first to get a proper rational expression before decomposing.

Write down the degree of each polynomial and confirm the numerator degree is smaller. If it’s not, stop and divide first.

Step 2: Factor the denominator completely

Now that you’ve confirmed the degree condition is satisfied, factor the denominator as far as possible. Break it into linear factors, repeated linear factors, and irreducible quadratic factors.

Write the fully factored form clearly because this determines what decomposition terms you’ll use in the next step. Every distinct factor and every repeated factor will contribute specific terms to your partial fraction setup.

Check that you can’t factor any term further. If you can, keep factoring until you reach terms that are either linear (ax + b) or irreducible quadratics (ax² + bx + c that don’t factor over the reals).

Step 3: Assign partial fraction terms based on factors

With the denominator fully factored, write the partial fractions decomposition using these rules based on factor type.

For a simple linear factor (ax + b) that appears once: use the term A/(ax + b).

For a repeated linear factor (ax + b)^k: use a sum of terms A₁/(ax + b) + A₂/(ax + b)² + … + A_k/(ax + b)^k.

For a simple irreducible quadratic (ax² + bx + c): use the term (Ax + B)/(ax² + bx + c).

For a repeated irreducible quadratic (ax² + bx + c)^k: use a sum of terms (A₁x + B₁)/(ax² + bx + c) + (A₂x + B₂)/(ax² + bx + c)² + … + (A_kx + B_k)/(ax² + bx + c)^k.

Write down all the terms you need based on these patterns. Each factor contributes specific terms, and missing a term or using the wrong form means the decomposition won’t work.

Step 4: Combine the partial fractions back into one expression

Now multiply each partial fraction term by the appropriate factors to create a common denominator that matches the original fully factored denominator.

For each term, multiply numerator and denominator by whatever factors are missing. When you combine all terms over the common denominator, the denominator should match exactly what you started with.

Write the combined numerator clearly because this is what you’ll use to solve for the unknown constants.

Step 5: Set numerators equal and solve for constants

Ignore the denominators since they’re equal on both sides, and set the original numerator equal to the combined numerator from your partial fraction decomposition.

Choose specific values of x that will eliminate all but one unknown at a time. If you have a factor (x – a), plug in x = a to make that factor zero and eliminate any term containing it. This isolates one constant so you can solve directly.

Work through enough x values to solve for all unknown constants. Each strategic x choice should eliminate enough terms to isolate one unknown cleanly.

Write down each constant as you solve for it.

Step 6: Integrate the decomposed terms

With all constants solved, substitute them back into your partial fraction decomposition. Now you have a sum of simple fractions that integrate using standard rules.

For terms like A/(ax + b), the integral is (A/a)ln|ax + b|.

For terms like A/(ax + b)ⁿ where n > 1, use the power rule for integration.

For terms like (Ax + B)/(ax² + bx + c), you may need to complete the square or split into two integrals.

Integrate each term separately, then combine the results and add the constant of integration.

These steps turn any partial fractions problem into a mechanical process. Once you’ve factored and assigned terms correctly, the rest is algebra and standard integration—no guessing required.

Example Problem: Partial Fractions

The workflow gives you the structure to decompose any rational expression methodically—check degrees, factor completely, assign terms by type, solve for constants strategically. Now let’s apply these steps to a real problem so you see exactly how the setup flows and where the checkpoints matter most.

With that laid out, let’s put these steps into practice.

This problem states:

Solution: Partial Fractions

When you see a factored denominator like (x + 3)(x – 2) and you’ve set up A/(x + 3) + B/(x – 2), the question isn’t whether you can combine fractions—it’s which x values to plug in that will eliminate terms cleanly so you solve for one constant at a time.

Plug in x = 0 out of habit, and both terms stay alive, forcing you into simultaneous equations.

Plug in x = 2 or x = -3, and entire terms vanish, giving you each constant directly.

That choice—strategic x values versus arbitrary ones—is the difference between thirty seconds of clean arithmetic and two minutes of equation manipulation that might introduce errors.

This is exactly why the partial fractions workflow matters. It turns the constant-solving step from “try numbers and see what happens” into “plug in values that make specific factors zero.” No guessing. No simultaneous equations unless absolutely necessary.

We’re going to walk it out step by step, the same way you would across the table, starting with the degree check and building through to the final integrated form.

Step 1: Check the degree condition

The first thing we need to do is confirm that partial fractions decomposition is valid for this expression.

The numerator is 5x, which is a first-degree polynomial (degree 1).

The denominator is x² + x – 6, which is a second-degree polynomial (degree 2).

Since the numerator degree (1) is less than the denominator degree (2), we can proceed directly with partial fractions decomposition. No polynomial division is needed.

Step 2: Factor the denominator completely

Next, we need to factor the denominator x² + x – 6 as completely as possible.

Looking for two numbers that multiply to -6 and add to +1, we find +3 and -2.

The factored form is:

x² + x – 6 = (x + 3)(x – 2)

Both factors are linear and cannot be factored further, so we’re done factoring.

Step 3: Assign partial fractions terms based on factors

When we have two distinct linear factors (x + 3) and (x – 2), the partial fractions decomposition takes the form:

5x / [(x + 3)(x – 2)] = A/(x + 3) + B/(x – 2)

We have two unknowns, A and B, that we need to solve for.

Step 4: Combine the partial fractions back into one expression

To combine the right side over a common denominator, we multiply each term by the opposite factor:

A/(x + 3) · (x – 2)/(x – 2) + B/(x – 2) · (x + 3)/(x + 3)

This gives us:

[A(x – 2) + B(x + 3)] / [(x + 3)(x – 2)]

The denominator now matches our original factored denominator, so we’re ready to solve for the constants.

Step 5: Set numerators equal and solve for constants

Since the denominators are equal, the numerators must be equal:

5x = A(x – 2) + B(x + 3)

Now we choose x values strategically to eliminate terms and solve for one constant at a time.

Solving for A: Choose x = -3 to make the B term zero:

5(-3) = A(-3 – 2) + B(-3 + 3)

-15 = A(-5) + 0

-15 = -5A

A = 3

Solving for B: Choose x = 2 to make the A term zero:

5(2) = A(2 – 2) + B(2 + 3)

10 = 0 + 5B

10 = 5B

B = 2

Step 6: Integrate the decomposed terms

Now our partial fractions decomposition is:

5x / [(x + 3)(x – 2)] = 3/(x + 3) + 2/(x – 2)

Integrating both sides:

∫ [3/(x + 3) + 2/(x – 2)] dx

= 3ln|x + 3| + 2ln|x – 2| + C

Rearranging to match answer format:

= 2ln|x – 2| + 3ln|x + 3| + C

So the final answer to this problem is A) 2ln|x – 2| + 3ln|x + 3| + C.

This tells us that the rational expression 5x/(x² + x – 6) breaks into two simple logarithmic terms that integrate directly once the partial fractions decomposition assigns the correct constants to each factor.

Common Mistakes to Avoid on Partial Fractions Problems

You know how to factor denominators—that’s algebra you learned years ago. Partial fractions problems fail in the ten seconds after factoring, when you see (x + 3)(x – 2) and write down the decomposition setup.

Write A/(x + 3) + B/(x – 2) for these two linear factors and you’re fine. But if that denominator were (x + 3)²(x – 2), and you still write just two terms instead of three, you’ve created a system with two unknowns when you actually need three. Your constants won’t solve. The numerators won’t balance. And you’ll waste two minutes checking algebra that was never going to work because the setup was broken from the start.

Here’s where execution breaks down on partial fractions and how to protect against it.

Mistake 1: Forgetting to check the degree condition first

Students see a rational expression and immediately start factoring the denominator without checking whether partial fractions even applies.

If the numerator degree is greater than or equal to the denominator degree, you cannot decompose directly. You must perform polynomial long division first to separate the expression into a polynomial part plus a proper rational expression.

On a partial fractions problem like our example with 5x/(x² + x – 6), the numerator degree (1) is less than denominator degree (2), so decomposition works directly. But if the numerator were x² + 5x, you’d need division first.

Skipping this check means you’ll set up a decomposition that can’t possibly work, and you won’t realize the error until your constants don’t solve correctly.

Mistake 2: Using the wrong decomposition terms for repeated factors

When a factor appears more than once in the denominator—like (x + 3)²—students write down A/(x + 3) and move on, treating it like a simple linear factor.

A repeated factor (x + a)^k requires k terms in the decomposition: A₁/(x + a) + A₂/(x + a)² + … + A_k/(x + a)^k.

So if our example denominator were (x + 3)²(x – 2) instead of just (x + 3)(x – 2), you’d need three terms total:

• A/(x + 3) for the first power
• B/(x + 3)² for the second power
• C/(x – 2) for the distinct linear factor

Missing terms means you don’t have enough unknowns to match the numerator structure, and the equation won’t balance.

This mistake breaks the setup immediately, but you won’t notice until you try to solve for constants and realize the algebra doesn’t close.

Mistake 3: Choosing x values that create complicated algebra

After combining the partial fractions and setting numerators equal, students plug in x = 0 or x = 1 out of habit, creating messy equations with multiple unknowns still present.

In our example, after setting 5x = A(x – 2) + B(x + 3), plugging in x = 0 gives:

0 = A(-2) + B(3), which is -2A + 3B = 0

You now have one equation with two unknowns. You need another equation, which means more work.

The point of choosing specific x values is to make entire terms disappear so you solve for one constant at a time. Choosing x = -3 makes the B(x + 3) term zero: 5(-3) = A(-3 – 2), giving you A directly.

Mistake 4: Losing track of which terms get which denominators when combining

When multiplying through to create a common denominator, students mix up which factors multiply with which terms, creating a combined numerator that doesn’t actually match the original denominator structure.

In our example with A/(x + 3) + B/(x – 2):

• The A term needs to be multiplied by (x – 2) to get full denominator
• The B term needs to be multiplied by (x + 3) to get full denominator

If you accidentally multiply A by (x + 3) instead, your combined form will be wrong, and when you set numerators equal, the constants won’t solve correctly.

Mistake 5: Forgetting that irreducible quadratics need (Ax + B) in the numerator

When the denominator contains a quadratic that doesn’t factor—like x² + 4—students write the decomposition term as A/(x² + 4), using only a constant in the numerator.

Irreducible quadratic factors require a linear numerator: (Ax + B)/(x² + 4). Using just a constant gives you one unknown when you need two to match the numerator structure properly.

If our example had an irreducible quadratic—say the denominator factored to (x + 3)(x² + 4)—your decomposition would be:

A/(x + 3) + (Bx + C)/(x² + 4)

Note the linear numerator (Bx + C) on the quadratic term, not just a constant.

This mistake means you don’t have enough degrees of freedom to solve the system, and the constants won’t balance correctly.

Rules of Thumb for Partial Fractions Problems on the FE

The workflow gives you the structure to match factors to decomposition terms correctly: (x – 2) gets A/(x – 2), (x + 3)² gets B/(x + 3) + C/(x + 3)², irreducible quadratics get linear numerators. These checkpoints protect you from writing the wrong number of terms, choosing x values that don’t eliminate anything, or setting up integrals you can’t evaluate because the decomposition was incomplete.

• Degree check comes first, always: Before you write down any partial fractions decomposition terms, confirm the numerator degree is strictly less than the denominator degree. If it’s not, you need polynomial division before decomposition. Skipping this costs you the entire problem because the decomposition won’t work.
• Match your decomposition terms exactly to the factor structure: Every distinct linear factor (ax + b) gets one term A/(ax + b). Every repeated factor (ax + b)^k gets k terms with increasing powers. Every irreducible quadratic gets (Ax + B) in the numerator. Your partial fractions decomposition term count must match your factor structure—if it doesn’t, the system won’t solve.
• Strategic x values eliminate terms cleanly: When solving for constants, choose x values that make specific factors zero. If you have (x – 3), plug in x = 3. If you have (x + 5), plug in x = -5. This eliminates entire terms and isolates one unknown at a time, turning simultaneous equations into direct calculations.
• Verify your combined denominator before solving: After multiplying through to combine partial fractions, stop and check that your combined denominator matches the original factored form exactly. If they don’t match, you multiplied wrong, and solving for constants will fail even if the algebra looks clean.
• Linear factors integrate to natural logs, powers integrate using the power rule: Once you have your partial fractions decomposition, terms like A/(x + b) integrate to (A)ln|x + b|. Terms like A/(x + b)ⁿ integrate using the power rule for n > 1. If you completed the decomposition correctly, every term should integrate using a standard rule—no complex substitutions needed.
• Don’t forget the constant of integration: After integrating all the decomposed terms, add + C at the end. It’s easy to drop when you’re focused on combining multiple logarithm terms, but it’s required in every indefinite integral.

Partial fractions is mechanical once the setup is right. These checkpoints keep the structure clean so you’re solving straightforward algebra instead of untangling mistakes you made three steps earlier.

Final Thoughts | Partial Fractions

Every partial fractions problem hinges on one decision: matching your factored denominator to the correct decomposition terms.

Get (x + 3)(x – 2) matched to A/(x + 3) + B/(x – 2), and the rest is arithmetic—combine fractions, set numerators equal, solve for constants, integrate.

Miss a term because you forgot the repeated factor rule, and your constants won’t solve no matter how perfect your algebra is.

Write B/(x + 3)² when you actually needed B/(x + 3) + C/(x + 3)², and you’ll burn two minutes before realizing the setup was wrong from step one.

The partial fractions workflow removes that fragility. It gives you explicit factor-to-term rules you execute before touching any algebra: simple linear factors get one term, repeated factors get multiple terms with increasing powers, irreducible quadratics get linear numerators. You’re not relying on memory or pattern recognition. You’re following mechanical rules that work the same way whether the denominator has two factors or six.

That’s what makes partial fractions manageable on the FE. Not memorizing decomposition patterns. Not hoping you remember which term goes where. Just matching factor types to term structures systematically, then executing the algebra you already know.

You’ve now got that systematic approach. The factor-to-term rules. The strategic x-value selection. The verification checkpoints that catch setup errors before they cascade.

Practice until you can look at a factored denominator and write the complete decomposition setup correctly on the first try—no erasing, no second-guessing, just clean execution from factors to terms to constants to integration. That’s when you know the partial fractions method is locked in.

Want more practice? Check out our complete FE problem library here.

You’re three weeks out from exam day and you still freeze when you see (x + 2)³ in a denominator because you can’t remember: is that one term or three?

Do I need A/(x + 2)³ only, or A₁/(x + 2) + A₂/(x + 2)² + A₃/(x + 2)³? That split-second hesitation costs you the problem, and it keeps happening because you’re grinding random integration problems hoping the decomposition rules stick instead of drilling the specific factor-to-term patterns that actually show up on the FE.

Every hour you spend practicing without a clear system for partial fractions is an hour you can’t get back.

Prepineer gives you targeted practice on the exact decomposition scenarios the FE tests—repeated linear factors, irreducible quadratics, mixed factor types—with step-by-step breakdowns that show you exactly where your setup breaks and how to fix it before the algebra even starts. You’re not guessing your way through decompositions. You’re building systematic execution with coaching support when the factor matching gets unclear.

Start your free 7-day trial and turn partial fractions from a method you’re uncertain about into a process you can execute confidently under pressure.