Integration by Parts Explained with Examples
Master integration by parts with the LIATE rule, tabular method, and worked examples including ln, inverse trig, and repeated integration.
Integration by Parts Explained
Integration by parts is one of the most important techniques in calculus. It lets you break down integrals that involve products of functions — things like xe^x, xsin(x), or ln(x) — into simpler pieces. Once you understand the formula and learn how to pick u and dv, most of these problems become straightforward.
This guide covers the formula, the LIATE rule for choosing u, the tabular method for repeated application, five worked examples, and the mistakes that trip students up most often. For a broader view of calculus topics, see our Calculus Guide.
The Formula
Integration by parts comes from the product rule for derivatives, run in reverse:
integral of u dv = uv - integral of v du
That is the entire formula. The challenge is never the formula itself — it is choosing which part of your integrand to call u and which to call dv.
The LIATE Rule
LIATE is a priority list for choosing u. Pick u as the function type that appears earliest in this list:
- L — Logarithmic functions (ln(x), log(x))
- I — Inverse trig functions (arcsin(x), arctan(x))
- A — Algebraic functions (x, x², polynomials)
- T — Trigonometric functions (sin(x), cos(x))
- E — Exponential functions (e^x, 2^x)
The idea: u should be something that gets simpler when you differentiate it, and dv should be something you can easily integrate.
LIATE is a guideline, not a law. It works for the vast majority of textbook problems.
Example 1: integral of x*e^x dx
By LIATE, u = x (algebraic) and dv = e^x dx.
- du = dx
- v = e^x
Apply the formula:
integral of xe^x dx = xe^x - integral of e^x dx = x*e^x - e^x + C = e^x(x - 1) + C
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Example 2: integral of ln(x) dx
This one surprises students because it does not look like a product. The trick: set u = ln(x) and dv = dx.
- du = (1/x) dx
- v = x
integral of ln(x) dx = xln(x) - integral of x(1/x) dx = xln(x) - integral of 1 dx = **xln(x) - x + C**
Example 3: integral of x*sin(x) dx
LIATE says u = x, dv = sin(x) dx.
- du = dx
- v = -cos(x)
integral of xsin(x) dx = -xcos(x) - integral of -cos(x) dx = -x*cos(x) + sin(x) + C
Final answer: sin(x) - x*cos(x) + C
Example 4: integral of arctan(x) dx (Inverse Trig)
Set u = arctan(x), dv = dx.
- du = 1/(1 + x²) dx
- v = x
integral of arctan(x) dx = x*arctan(x) - integral of x/(1 + x²) dx
The remaining integral is a simple substitution (let w = 1 + x², dw = 2x dx):
= x*arctan(x) - (1/2)ln(1 + x²) + C
Final answer: x*arctan(x) - (1/2)ln(1 + x²) + C
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Example 5: Repeated Integration by Parts — integral of x²*e^x dx
Sometimes one round is not enough. Here u = x² and dv = e^x dx.
Round 1:
- u = x², du = 2x dx, v = e^x
- = x²e^x - integral of 2xe^x dx
Round 2 (on the remaining integral):
- u = 2x, du = 2 dx, v = e^x
- = x²e^x - [2xe^x - integral of 2*e^x dx]
- = x²e^x - 2xe^x + 2e^x + C
Final answer: e^x(x² - 2x + 2) + C
This is where the tabular method shines.
The Tabular Method
When you need to apply integration by parts multiple times — typically when u is a polynomial — the tabular method saves time and reduces errors.
Set up a table with three columns: alternating signs, derivatives of u, and integrals of dv.
For integral of x²*e^x dx:
| Sign | Derivatives of x² | Integrals of e^x |
|---|---|---|
| + | x² | e^x |
| - | 2x | e^x |
| + | 2 | e^x |
| - | 0 | e^x |
Multiply diagonally and apply the signs:
(+)x²e^x + (-)2xe^x + (+)2*e^x = e^x(x² - 2x + 2) + C
Same answer, far less writing. Use the tabular method whenever u is a polynomial multiplied by an exponential or trig function.
Tabular method problems can get long. Let Math.Photos handle the bookkeeping — screenshot the integral and get every row of the table shown clearly.
Example 6: The “Boomerang” — integral of e^x*sin(x) dx
Some integrals require integration by parts twice, then solving for the original integral algebraically.
Let I = integral of e^x*sin(x) dx. Set u = sin(x), dv = e^x dx.
Round 1: I = e^xsin(x) - integral of e^xcos(x) dx
Round 2 (u = cos(x), dv = e^x dx):
I = e^xsin(x) - [e^xcos(x) + integral of e^x*sin(x) dx]
I = e^xsin(x) - e^xcos(x) - I
Add I to both sides:
2I = e^x(sin(x) - cos(x))
I = e^x(sin(x) - cos(x))/2 + C
This “boomerang” technique is essential for integrals combining exponentials with trig functions.
Common Mistakes
1. Wrong choice of u and dv. If your integral gets harder after applying the formula, you probably chose u and dv backwards. Try swapping them.
2. Sign errors. The formula has a subtraction. Combined with negative signs from trig integrals, it is easy to drop or flip a sign. Write every step.
3. Forgetting + C. Indefinite integrals always need the constant of integration.
4. Not recognizing when to stop. After applying integration by parts, the remaining integral should be simpler. If it is not, reconsider your approach — you may need a different technique entirely.
5. Boomerang problems: dividing wrong. When the original integral reappears, make sure you move it to the correct side before dividing. A sign mistake here ruins the whole solution.
FAQ
When should I use integration by parts versus substitution?
Use substitution when you spot an inner function and its derivative nearby (like xcos(x²), where 2x is the derivative of x²). Use integration by parts when the integrand is a product of two different types of functions (like xcos(x), a polynomial times trig) and no clean substitution exists.
What if I need to apply integration by parts more than twice?
Use the tabular method. It handles any number of rounds cleanly as long as one factor eventually differentiates to zero (polynomials always do). For boomerang-style integrals where neither factor vanishes, two rounds plus algebra is the standard approach.
Does integration by parts work for definite integrals?
Yes. The formula becomes: integral from a to b of u dv = [uv] from a to b minus integral from a to b of v du. Evaluate the uv term at both bounds. Everything else works the same way.
How can I verify my answer?
Differentiate your result. If you get back the original integrand, the integration is correct. You can also use Math.Photos to check your work — screenshot the problem and compare solutions side by side.
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