Integration by Parts: How to Integrate Products of Functions Without Guessing

The integral ∫₀¹ x·e^x dx takes thirty seconds if you trust the LIATE priority system and know that x becomes u while e^x becomes dv.

It takes three minutes if you reverse that choice, differentiate the exponential first, integrate the polynomial, and end up applying integration by parts multiple times to chase down an answer that should have been straightforward.

Integration by parts exists because the product rule exists, and the choice you make—which function to differentiate versus which to integrate—determines whether you simplify the problem or expand it into recursive applications that burn time and create more opportunities for algebraic errors.

This guide walks you through a systematic approach that works on any integration by parts problem the FE throws at you—from straightforward polynomial-exponential products to nested combinations requiring multiple applications.

You’ll learn exactly how to choose u and dv using LIATE priority without trial and error, how to execute the formula cleanly, and how to verify your setup simplifies rather than complicates the remaining integral.

Before we walk through it step by step, watch this short video.

It shows you the full process from identifying product structure to applying the LIATE system to executing the formula without sign errors. You’ll see exactly where students lose control when they reverse the u and dv choice and how to verify the remaining integral is simpler before committing to the algebra.

What Is Integration by Parts?

Integration by parts is the integration counterpart to the product rule for differentiation.

When you have two functions multiplied together and you need to integrate that product, you can’t just integrate each piece separately and multiply the results—that violates the fundamental theorem. The integration by parts formula gives you a method that transforms the product integral into something you can actually solve.

The technique works by differentiating one function (making it simpler) while integrating the other (which you handle separately), then combining them according to a specific formula.

What you’re really doing is undoing the product rule in reverse—you’re recognizing that your integral came from a product rule derivative, and you’re working backward through that process.

Think of it like translating a complex sentence. You can’t translate word by word and get the right meaning.

You have to understand the structure first, break it into pieces that make sense, then reconstruct it in a way that works. Integration by parts is that structural translation for products.

It matters because product structures show up constantly in engineering—any time you’re modeling decay with time weights, analyzing moment calculations with distance functions, or working with Laplace transforms in controls and circuits.

The FE doesn’t ask you to derive the technique. It asks you to execute it correctly under time pressure when the setup isn’t obvious.

On the FE Exam, integration by parts shows up when you see products like x·e^x, x²·sin(x), or ln(x)·anything, and the problem asks you to evaluate a definite integral or find an antiderivative.

The key is recognizing the pattern fast and choosing your split correctly the first time.

How to Work Through Integration by Parts Problems

When you see x multiplied by e^x, you know you need integration by parts—the product rule won’t help you going forward, and you can’t integrate each piece separately and multiply.

The breakdown happens when you’re deciding which function becomes u (the one you differentiate) and which becomes dv (the one you integrate).

Choose wrong—make e^x your u and x your dv—and you’ll differentiate the exponential into e^x, integrate the polynomial into x²/2, and face ∫ (x²/2)·e^x dx.

That’s worse than where you started. The remaining integral is more complex, requires integration by parts again, and every additional application multiplies your chances of dropping a term or flipping a sign.

LIATE priority removes that guesswork. Algebraic comes before Exponential in the priority order, so x becomes u and e^x dx becomes dv.

Differentiate x to get du = dx. Integrate e^x to get v = e^x.

The remaining integral ∫ e^x dx is simpler than the original—no recursion needed, just one straightforward integration.

Let’s lay out the steps.

Step 1: Identify the Product Structure and Assign Functions

The first thing you need to do is confirm you’re looking at a product—two functions multiplied together inside an integral, not a single composite function. Write the integral clearly and identify both pieces.

For ∫ x·e^x dx, you have x (a polynomial) multiplied by e^x (an exponential). That’s a product, and integration by parts is the technique that handles it.

Once you confirm it’s a product, your job is to decide which function becomes u (the one you’ll differentiate) and which becomes dv (the one you’ll integrate). This isn’t arbitrary. Use the LIATE priority system:

• Logarithmic functions (ln(x), log(x))
• Inverse trig functions (arcsin(x), arctan(x))
• Algebraic functions (polynomials: x, x², x³)
• Trigonometric functions (sin(x), cos(x))
• Exponential functions (e^x, e^(kx))

Choose u as the function that appears earlier in LIATE. Choose dv as the function that appears later.

This priority works because functions higher on the list simplify when differentiated, and functions lower on the list integrate cleanly without complicating.

For x·e^x: Algebraic (x) comes before Exponential (e^x) in LIATE. So u = x and dv = e^x dx.

Step 2: Compute du and v

Now that you’ve assigned u and dv, differentiate u to get du, and integrate dv to get v. Write both clearly before moving forward.

For u = x, differentiate: du = dx. The derivative of x is 1, and the 1·dx simplifies to dx.

For dv = e^x dx, integrate: v = e^x. The exponential function e^x is special—it’s its own derivative and its own integral.

Don’t add the constant of integration here—you’ll handle that at the end when you’ve completed the full integral.

Step 3: Apply the Integration by Parts Formula and Evaluate the Remaining Integral

With u, v, du, and dv clearly defined, plug them into the integration by parts formula: ∫ u dv = uv – ∫ v du.

Write the formula out explicitly first:

∫ u dv = uv – ∫ v du

Now substitute your values:

∫ x·e^x dx = x·e^x – ∫ e^x dx

The first term uv = x·e^x is a product. You’re done with that piece—it stays as is.

The second term is the remaining integral ∫ e^x dx. This is straightforward—the integral of e^x is e^x:

= x·e^x – e^x

Factor out e^x:

= e^x(x – 1)

For a definite integral from 0 to 1, you’d now apply the limits of integration to this entire result. Evaluate at the upper limit (1), evaluate at the lower limit (0), and subtract:

[e^x(x – 1)]₀¹ = [e¹(1 – 1)] – [e⁰(0 – 1)]

= [e·0] – [1·(-1)]

= 0 – (-1)

= 1

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

Example Problem: Integration by Parts

The workflow handles any integration by parts problem once you internalize the LIATE selection system.

The goal is to turn product integrals into clean setups that simplify predictably, so you’re not guessing or backtracking on exam day.

Right now, we’re going to work through an FE-style problem to see exactly where the LIATE choice protects you and how the algebra flows once you’ve committed to u and dv correctly.

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

This problem states:

Solution: Integration by Parts

When you see x multiplied by e^x, the question isn’t whether you need integration by parts—you do, because this is a product and the product rule doesn’t help you integrate.

The question is whether x or e^x should be u, and whether making the wrong choice means you’ll expand the problem into something that requires two or three applications instead of one.

That decision happens in the first ten seconds.

Choose wrong, and you won’t know it until you’ve already applied the formula and the remaining integral is more complicated than what you started with.

The workflow removes that uncertainty by giving you LIATE priority: Algebraic before Exponential means x becomes u, e^x dx becomes dv, and the remaining integral simplifies instead of expanding.

Let’s walk it out step by step, exactly how you’d execute it on the FE.

Step 1: Identify the Product and Assign u and dv Using LIATE

The first thing we need to do is read through the problem and confirm we’re looking at a product structure that requires integration by parts.

We have x (algebraic) multiplied by e^x (exponential). That’s a product. Now assign u and dv using LIATE priority.

Algebraic functions come before Exponential in LIATE, so:

• u = x
• dv = e^x dx

Differentiate u to get du:

• du = dx

Integrate dv to get v:

• v = e^x

Step 2: Apply the Integration by Parts Formula

Next, we need to plug our values into the integration by parts formula: ∫ u dv = uv – ∫ v du.

Substituting:

∫ x·e^x dx = x·e^x – ∫ e^x dx

Now we need to evaluate the remaining integral ∫ e^x dx.

The integral of e^x is e^x, so:

= x·e^x – e^x

Factor out e^x:

= e^x(x – 1)

That’s our antiderivative. Now we apply the definite integral limits from 0 to 1.

Step 3: Apply Limits of Integration and Calculate Final Answer

When we evaluate the definite integral ∫₀¹ x·e^x dx, we take our combined result and substitute the upper and lower bounds.

Our antiderivative is: e^x(x – 1)

Evaluate at x = 1:

e¹(1 – 1) = e(0) = 0

Evaluate at x = 0:

e⁰(0 – 1) = 1(-1) = -1

Subtract (upper limit minus lower limit):

0 – (-1) = 0 + 1 = 1

Final answer: 1.000

The final answer is B) 1.000.

This tells us that the area under the curve x·e^x from 0 to 1 equals exactly 1 square unit. The polynomial term x weights the exponential growth across the interval, and when integrated by parts and evaluated at the boundaries, the contributions from both the uv term and the remaining integral combine to yield a clean integer result.

Common Mistakes to Avoid on Integration by Parts Problems

Integration by parts problems with polynomial-exponential products break at the moment you assign u and dv.

Choose e^x as u instead of x, and you’ll differentiate the exponential (getting e^x again), integrate the polynomial (getting x²/2), and face ∫ (x²/2)·e^x dx—which requires integration by parts twice more instead of once.

That choice cascades. Every additional application creates more terms to track, more opportunities for sign errors, and more algebra where dropping a coefficient or reversing a limit turns a correct setup into a wrong answer.

Mistake 1: Choosing u and dv Backward

This happens when you see x·e^x and you make e^x your u because exponentials “look harder” or you think they need to be differentiated.

You differentiate e^x and get e^x (it doesn’t simplify). You integrate x and get x²/2. Now your remaining integral is ∫ (x²/2)·e^x dx.

You’ve made the problem worse. The polynomial degree increased from 1 to 2.

You need to apply integration by parts again—this time with u = x² and dv = e^x dx—and then again after that because you’ll still have ∫ 2x·e^x dx remaining.

What should have taken one application now takes three. Each additional step multiplies opportunities for sign errors, dropped terms, or coefficient mistakes.

Mistake 2: Forgetting the Negative Sign in the Formula

The integration by parts formula is ∫ u dv = uv – ∫ v du. That minus sign isn’t optional—it’s structural.

When you write x·e^x and then add ∫ e^x dx instead of subtracting it, your final answer doubles the contribution of the second term.

For ∫₀¹ x·e^x dx, if you write x·e^x + e^x instead of x·e^x – e^x, you get e^x(x + 1) as your antiderivative.

When you evaluate from 0 to 1, you’ll calculate [e·2 – 1·1] = 2e – 1 ≈ 4.44 instead of 1. That doesn’t match any answer choice, and you’ll burn minutes hunting for an error that lives in a single dropped sign.

Mistake 3: Applying Limits of Integration Before Completing the Integral

When you’re working the definite integral ∫₀¹ x·e^x dx, you apply integration by parts and get x·e^x – ∫ e^x dx.

The instinct is to apply the limits [0 to 1] immediately to the first term: evaluate x·e^x at 1, evaluate it at 0, subtract, then deal with the remaining integral ∫ e^x dx separately.

That creates errors. You end up applying limits twice—once to uv and once to the integral result—and you lose track of whether you’ve subtracted correctly or double-counted boundary terms.

Your final answer will be off, usually landing on answer choice C) 1.718 (which is e – 1, the result when you evaluate x·e^x from 0 to 1 but forget to subtract the remaining integral).

Mistake 4: Evaluating the Antiderivative Incorrectly at the Bounds

When you have e^x(x – 1) and you need to evaluate it from 0 to 1, the algebra at each bound matters.

At x = 1, you get e¹(1 – 1) = e(0) = 0. At x = 0, you get e⁰(0 – 1) = 1(-1) = -1.

If you rush and write e⁰ = 0 (forgetting that any number to the zero power is 1), you’ll calculate 0 – 0 = 0 instead of 0 – (-1) = 1.

Or if you evaluate (x – 1) at x = 1 and get confused by the zero, you might think the whole term vanishes and just evaluate at x = 0, giving you -1 as your final answer.

Rules of Thumb for Integration by Parts Problems on the FE

LIATE priority handled the x·e^x split correctly: Algebraic before Exponential meant u = x and dv = e^x dx.

Differentiating x gave du = dx, which is simpler. Integrating e^x gave v = e^x, which stayed clean.

The remaining integral ∫ e^x dx resolved in one step instead of requiring recursion.

These checkpoints protect that process when you’re moving fast and the structure isn’t immediately obvious.

• LIATE priority isn’t negotiable: On the FE, you don’t have time to test multiple splits and see which one works. Use LIATE every time: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Choose u as the function earlier in that order, dv as the function later. For x·e^x, that means u = x (Algebraic) and dv = e^x dx (Exponential). This prevents the most common setup error—reversing the choice and making the problem require multiple applications.
• Write the formula before substituting values: Don’t try to apply integration by parts mentally or skip straight to plugging in u, v, du, and dv. Write ∫ u dv = uv – ∫ v du explicitly on your page, then substitute. This forces you to see the structure and prevents dropping the minus sign or mixing up terms when the algebra gets dense.
• For definite integrals, integrate first and apply limits last: Complete all integration by parts steps, combine uv and the result of ∫ v du into a single antiderivative, and only then apply the upper and lower limits of integration. For ∫₀¹ x·e^x dx, that means finishing e^x(x – 1) before substituting x = 1 and x = 0. Applying limits in the middle creates boundary term errors and sign mistakes that are hard to trace back and fix under time pressure.
• Check that ∫ v du is simpler before proceeding: After choosing u and dv, compute du and v, then look at the remaining integral ∫ v du. If it’s more complicated than your original integral, stop—you’ve made the problem worse. For x·e^x, checking ∫ e^x dx confirms it’s simpler than ∫ x·e^x dx because the polynomial term is gone. If that check fails, reverse your choice of u and dv and start over.
• Factor the antiderivative before applying limits: After completing integration by parts, look for common factors you can pull out. For x·e^x – e^x, factoring gives e^x(x – 1), which is cleaner to evaluate at bounds. When x = 1, the (x – 1) term becomes zero immediately, and you don’t waste time computing e¹·0. When x = 0, you get 1·(-1) = -1 directly. Factoring simplifies boundary evaluation and reduces arithmetic errors.

Integration by parts isn’t a guessing game. The formula is fixed, the LIATE priority system is reliable, and the algebra is straightforward if you keep the structure clean and apply limits only after completing all integration work.

Final Thoughts | Integration by Parts

Every integration by parts problem hinges on the LIATE split.

Get u and dv right—choose x as u and e^x dx as dv—and the remaining integral simplifies to ∫ e^x dx, which you finish in one step.

Reverse that choice—make e^x your u and x your dv—and you’ll face ∫ (x²/2)·e^x dx, which requires two more applications to resolve.

That decision happens in the first ten seconds of the problem.

You either trust LIATE priority and move forward with confidence, or you guess based on what looks simpler in the moment and discover three steps later that you’ve made the problem exponentially harder.

The difference between students who execute integration by parts cleanly and students who burn three minutes chasing recursive applications isn’t conceptual understanding.

It’s the habit of applying LIATE priority before writing anything else. Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential—earlier in that list becomes u, later becomes dv, every time.

On the FE, integration by parts problems are solvable in under two minutes if you use the priority system, write the formula before substituting, and don’t touch the limits until all integration work is complete.

That structure removes the guessing and turns product integrals into points you bank instead of time you bleed.

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