Integration by Parts: When to Use It and How It Works
You have learned the basic antiderivative rules. You have practiced u-substitution. But every now and then, a calculus problem shows up that neither of those methods can handle cleanly. Two different types of functions are multiplied together — and there is no obvious way to simplify things. That is the moment when integration by parts steps in.
Integration by parts is one of the most important techniques in all of calculus. It gives you a reliable way to integrate products of functions — things like x · eˣ or x · cos(x) — that would be impossible to solve with basic rules alone.
In this guide, you will learn exactly when to use integration by parts, how the formula works, how to choose your u and dv correctly, and how to apply the method step by step. Clear worked examples are included throughout so the process feels completely manageable by the time you finish.
What Is Integration by Parts?
Integration by parts is a technique used to integrate the product of two functions. It is derived from the product rule in differentiation — the same way u-substitution comes from the chain rule. In fact, integration by parts is simply the product rule run in reverse.
Here is the formula:
∫ u · dv = u · v − ∫ v · du
Where u and v are both functions of x, chosen strategically from the original integral.
Do not let the formula intimidate you. In plain English, it says: instead of integrating the product directly, split it into two parts, do some differentiation and integration separately, then combine the results using the formula. The steps are very mechanical once you know the process.
The key to making this formula work is choosing the right u and the right dv from the original integral. That choice determines how easy or hard the remaining integral becomes — and there is a reliable strategy for making that choice every time.
When Should You Use Integration by Parts?
Before diving into the steps, it helps to know when this method is actually called for. Here are the clearest signals that integration by parts is the right tool:
- The integral contains two different types of functions multiplied together — for example, a polynomial and an exponential, or a polynomial and a trig function.
- U-substitution does not work because the derivative of one part does not appear elsewhere in the integral.
- You see combinations like x · eˣ, x² · sin(x), x · ln(x), or x · cos(x).
A quick general rule: if the integral looks like (algebraic function) × (exponential, trig, or log function), integration by parts is almost certainly the right approach.
The LIATE Rule Explained Simply
The LIATE rule is a memory tool that tells you which part of the integral to assign as u. Here is what each letter stands for and why the order matters:
- L — Logarithms: Functions like ln(x). These are hard to integrate directly, so making them u means you only need to differentiate them — which is much easier.
- I — Inverse Trig: Functions like arctan(x) or arcsin(x). Same reason as logarithms — differentiation is simpler than integration for these.
- A — Algebraic: Polynomials like x, x², or 3x⁴. These simplify nicely when differentiated and are a natural middle choice.
- T — Trigonometric: Functions like sin(x) or cos(x). Easy to both differentiate and integrate, so they make good dv choices.
- E — Exponential: Functions like eˣ. These stay the same when integrated, making them excellent choices for dv.
Once you assign u, everything that remains in the integral — including dx — becomes dv. Then you differentiate u to get du, and integrate dv to get v. Those four pieces plug straight into the formula.
Step-by-Step Process for Integration by Parts
Here is the full method laid out in clear, repeatable steps:
- Identify that integration by parts is needed — two different function types multiplied together.
- Choose u and dv using the LIATE rule.
- Differentiate u to find du.
- Integrate dv to find v. Do not add + C at this stage.
- Plug everything into the formula: ∫ u · dv = u · v − ∫ v · du.
- Solve the remaining integral — ∫ v · du. This should now be simpler than the original.
- Simplify your answer, add + C, and verify by differentiating.
Notice step 4 says do not add + C when finding v. That is intentional — the + C only goes into the very final answer, not in the middle of the working. Adding it early creates unnecessary confusion without changing the result.
Worked Examples: Integration by Parts Step by Step
Step 1: Two functions multiplied together — x (algebraic) and eˣ (exponential). Integration by parts applies.
Step 2: Choose u and dv using LIATE.
Algebraic comes before Exponential in LIATE, so:
u = x and dv = eˣ dx
Step 3: Find du.
Differentiate u = x: du = dx
Step 4: Find v.
Integrate dv = eˣ dx: v = eˣ
Step 5: Apply the formula.
∫ x · eˣ dx = x · eˣ − ∫ eˣ · dx
Step 6: Solve the remaining integral.
∫ eˣ dx = eˣ
Step 7: Combine and add + C.
x · eˣ − eˣ + C = eˣ(x − 1) + C
✅ Answer: eˣ(x − 1) + C
Check: Differentiate using the product rule — eˣ(x − 1) + eˣ · 1 = eˣ · x − eˣ + eˣ = xeˣ. ✓
Step 1: Two functions — x (algebraic) and cos(x) (trig). Integration by parts is the right tool.
Step 2: Choose u and dv using LIATE.
Algebraic comes before Trigonometric, so:
u = x and dv = cos(x) dx
Step 3: Find du.
Differentiate u = x: du = dx
Step 4: Find v.
Integrate dv = cos(x) dx: v = sin(x)
Step 5: Apply the formula.
∫ x · cos(x) dx = x · sin(x) − ∫ sin(x) dx
Step 6: Solve the remaining integral.
∫ sin(x) dx = −cos(x)
Step 7: Combine and add + C.
x · sin(x) − (−cos(x)) + C = x · sin(x) + cos(x) + C
✅ Answer: x · sin(x) + cos(x) + C
Check: Differentiate using the product rule — sin(x) + x · cos(x) − sin(x) = x · cos(x). ✓
This one looks like it only has one function — but integration by parts still applies. The trick is to rewrite ln(x) as ln(x) · 1, and treat the 1 as the second function.
Step 1: Rewrite as ∫ ln(x) · 1 dx.
Step 2: Choose u and dv using LIATE.
Logarithm comes first in LIATE, so:
u = ln(x) and dv = 1 dx = dx
Step 3: Find du.
Differentiate u = ln(x): du = (1/x) dx
Step 4: Find v.
Integrate dv = dx: v = x
Step 5: Apply the formula.
∫ ln(x) dx = x · ln(x) − ∫ x · (1/x) dx = x · ln(x) − ∫ 1 dx
Step 6: Solve the remaining integral.
∫ 1 dx = x
Step 7: Combine and add + C.
✅ Answer: x · ln(x) − x + C
Check: Differentiate — ln(x) + x · (1/x) − 1 = ln(x) + 1 − 1 = ln(x). ✓
What If You Need to Apply Integration by Parts Twice?
Sometimes after applying the formula once, the remaining integral is still a product of two functions. In that case, you simply apply integration by parts a second time to that new integral. This is completely normal and comes up often with integrals like x² · eˣ or x² · cos(x).
The process is exactly the same the second time around — choose a new u and dv from the remaining integral, follow all seven steps again, and combine both results at the end.
A good habit is to keep your working clearly organized when applying the method twice. Label your first round of u, du, v, and dv, then clearly start a second round below. Keeping things tidy prevents small mix-ups that can throw off the whole answer.
Common Mistakes to Avoid
Integration by parts has several steps, which means there are several places where things can go wrong. Here are the mistakes beginners make most often:
- Choosing u and dv the wrong way around: If your remaining integral looks more complicated than the original, swap your choices. The whole point of the method is that ∫ v · du should be simpler than ∫ u · dv.
- Forgetting the minus sign in the formula: The formula is u · v minus ∫ v · du. Dropping or flipping that minus sign is one of the most common errors in all of integration.
- Adding + C when finding v: Save the + C for the very final answer only. Adding it when integrating dv to find v creates unnecessary complications mid-problem.
- Not simplifying ∫ v · du before moving on: Always simplify the remaining integral as much as possible before trying to solve it. A messy unsimplified integral leads to avoidable errors.
- Forgetting + C in the final answer: As with every antiderivative, the constant of integration must appear at the very end — every single time.
Frequently Asked Questions
How is integration by parts different from u-substitution?
U-substitution works when one function is nested inside another — a composite function — and its derivative appears elsewhere in the integral. Integration by parts works when two different functions are multiplied together and neither is the derivative of the other. If you see a product of two unrelated function types, reach for integration by parts. If you see a function inside another function, try u-substitution first.
What does LIATE stand for and do I always have to follow it?
LIATE stands for Logarithms, Inverse trig, Algebraic, Trigonometric, and Exponential. It is a guideline for choosing which part of the integral becomes u — always pick the type that appears earliest in that list. It works correctly for the overwhelming majority of beginner and intermediate problems. Occasionally an unusual integral may require breaking the rule, but when you are starting out, following LIATE consistently is the safest and most reliable approach.
Can I check my integration by parts answers with an antiderivative calculator?
Absolutely — and it is highly recommended. Integration by parts involves several steps, so errors can hide in the middle of the working rather than at the end. After solving a problem on your own, use the antiderivative calculator on this site to verify your final answer instantly. If the results do not match, go back through your steps one at a time — the mistake is usually a sign flip, a wrong choice of u, or a forgotten + C.
Conclusion
Integration by parts is a powerful and elegant technique that unlocks a whole new category of calculus problems. Remember the formula — ∫ u · dv = u · v − ∫ v · du — use the LIATE rule to make smart choices, follow the steps in order, and always verify your answer by differentiating at the end. With enough practice, the method becomes second nature and even complex-looking integrals start to feel approachable. Ready to test what you have learned? Use the antiderivative calculator on this site to check your integration by parts answers instantly and keep building your calculus confidence one problem at a time!
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