What if the new integral is as complicated as the first?

You likely assigned $u$ and $dv$ incorrectly. Swap your choices and try again. Alternatively, the problem might require repeated integration by parts to fully reduce the algebraic term.

How do I evaluate definite integrals by parts?

Apply the limits of integration to both the $uv$ boundary term and the new integral: $\int_a^b u \, dv = uv \Big|_a^b - \int_a^b v \, du$.

What is a repeating or circular integral?

When integrating products like $e^x \sin x$, applying parts twice returns a multiple of the original integral on the right side. You must treat the original integral as a variable, set up an algebraic equation, and solve for it.

How to integrate by parts

Integration by parts is a technique that transforms the integral of a product of functions into a simpler integral using the formula $\int u \, dv = uv - \int v \, du$ .

It applies primarily when the integrand is a product of algebraic and transcendental functions, or when standard $u$ -substitution fails.

The setup

Identify the integrand as a product of two factors, $u$ and $dv$ . Choose $u$ such that its derivative $du$ is simpler or of lower degree, and choose $dv$ such that its antiderivative $v$ is easily computable. The LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) is a standard hierarchy for prioritizing the choice of $u$ .

The steps

Assign $u$ and $dv$ from the given integrand.
Differentiate $u$ to find $du$ .
Integrate $dv$ to find $v$ . Do not add a constant of integration here.
Substitute $u$ , $v$ , $du$ , and $dv$ into the formula $\int u \, dv = uv - \int v \, du$ .
Evaluate the new integral $\int v \, du$ .
Add the constant of integration $C$ for indefinite integrals.

Checking the result

Differentiate the final answer with respect to the variable of integration. By applying the product rule to the $uv$ term, the result must simplify algebraically to exactly match the original integrand.

Common errors

Forgetting the negative sign in the integration by parts formula.
Dropping the constant of integration $C$ at the final step.
Choosing a $dv$ factor that lacks a known or simple antiderivative.
Creating an infinite loop by swapping the types of functions assigned to $u$ and $dv$ during a second application of parts.

Worked example

Evaluate $\int x e^x \, dx$ .

Let $u = x$ and $dv = e^x \, dx$ . Compute the differentials and integrals: $du = dx$ $v = e^x$ Apply the formula: $\int u \, dv = uv - \int v \, du$ $\int x e^x \, dx = x e^x - \int e^x \, dx$ Evaluate the remaining integral: $= x e^x - e^x + C$

FAQ

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Answer form

References: Stewart Calculus, 8th Edition, Chapter 7.1 · OpenStax Calculus Volume 2, Chapter 3.1 · Thomas' Calculus, 14th Edition, Chapter 8.2