Can I use integration by parts more than once in the same problem?

Yes, you often have to! If your new integral still contains a product of functions, you can apply the technique again. For example, integrating $x^2 e^x$ requires two passes to fully reduce the $x^2$ term to a constant.

What happens if I pick the wrong parts for u and dv?

If you pick the wrong parts, the new integral ($\int v du$) will look more complicated than the original one you started with. If you notice things getting messier, just stop, swap your choices for $u$ and $dv$, and start over.

Do I need to add the constant of integration (+ C) when I find v?

No, you can leave the constant of integration out during the intermediate steps. You only need to add $+ C$ at the very end of the entire problem after you evaluate the final integral.

When do you use integration by parts?

Integration by parts is used when you need to integrate the product of two different types of functions, like an algebraic term multiplied by a trigonometric term. It is essentially the reverse of the product rule for derivatives. Think of it like untangling two different types of threads woven together. You pick one thread to pull out and simplify (by differentiating it) and another to wind up (by integrating it), which makes the overall knot much easier to handle.

What is integration by parts?

Integration by parts is a technique that transforms a difficult integral of a product into a simpler one. The formula is $\int u dv = uv - \int v du$ . To use it, you split your original integrand into two parts: a part called $u$ that gets simpler when you take its derivative, and a part called $dv$ that is easy to integrate.

How to recognize when to use it

You should consider integration by parts when you see the product of unrelated function families. Common combinations include polynomials multiplied by exponential functions (like $x e^x$ ), polynomials multiplied by trigonometric functions (like $x \sin x$ ), or single functions that do not have basic integral formulas (like $\ln x$ or $\arctan x$ ). If standard substitution does not work on a product, integration by parts is usually your next best step.

How to choose your parts (The LIATE Rule)

Choosing the right $u$ is critical. A helpful mnemonic is LIATE: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential. You look at the two functions in your product and pick the one that comes first in the LIATE list to be your $u$ . The remaining part, including the $dx$ , becomes your $dv$ . This rule works because functions at the top of the list get much simpler when differentiated, while functions at the bottom are very easy to integrate repeatedly.

Where students slip

A common mistake is forgetting to include the $dx$ when setting up $dv$ , which throws off the notation for the rest of the problem. Another frequent error is messing up the negative sign in the $\int u dv = uv - \int v du$ formula, especially if the new integral also produces a negative term. Always use parentheses when subtracting the second integral to keep your signs straight.

Worked through

Evaluate the integral: $\int x e^x dx$

Using the LIATE rule, $x$ is Algebraic and $e^x$ is Exponential. Since Algebraic comes first, we set $u = x$ . That means $dv = e^x dx$ . Next, we find $du$ by differentiating $u$ , which gives $du = 1 dx$ . We find $v$ by integrating $dv$ , which gives $v = e^x$ . Now we plug these into the formula: $\int u dv = uv - \int v du$ . This gives us $x e^x - \int e^x dx$ . Finally, we evaluate the simpler integral: $\int e^x dx = e^x$ . Our final answer is $x e^x - e^x + C$ .

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Where this comes from: OpenStax Calculus Volume 2 · Khan Academy AP Calculus AB / BC · Stewart Calculus 8th Edition