Integration by parts is a method for integrals that contain a product of functions. It comes from the product rule for derivatives and is most useful when one part of the product becomes simpler after differentiation. The formula is \(\int u\,dv = uv - \int v\,du\). A good setup turns the original problem into an easier integral.
Why the formula works
The product rule says \((uv)' = u'v + uv'\), or in differential notation, \(d(uv)=u\,dv+v\,du\). Rearranging and integrating gives \(\int u\,dv = uv - \int v\,du\). This is why the method is not a trick; it is the product rule running backward.
Using LIATE
LIATE is a memory guide for choosing \(u\): logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. It is not a law, but it often works because logarithmic and inverse trigonometric functions usually simplify when differentiated, while exponential and trigonometric functions remain manageable when integrated.
Worked example 1: \(\int xe^x\,dx\)
Choose \(u=x\) because it becomes simpler when differentiated. Choose \(dv=e^x\,dx\) because it is easy to integrate. Then \(du=dx\) and \(v=e^x\). Apply the formula: \(\int xe^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x+C\). A quick derivative check returns \(xe^x\).
Worked example 2: \(\int x\cos x\,dx\)
Let \(u=x\) and \(dv=\cos x\,dx\). Then \(du=dx\) and \(v=\sin x\). The integral becomes \(x\sin x-\int \sin x\,dx\). Since \(\int \sin x\,dx=-\cos x\), the result is \(x\sin x+\cos x+C\).
Worked example 3: \(\int \ln x\,dx\)
This problem looks like it has one factor, but write it as \(\int 1\cdot \ln x\,dx\). Choose \(u=\ln x\) and \(dv=dx\). Then \(du=1/x\,dx\) and \(v=x\). The result is \(x\ln x-\int 1\,dx=x\ln x-x+C\).
Common mistakes
Students often lose the minus sign in the formula, choose a \(dv\) that cannot be integrated, or stop before simplifying the remaining integral. Another error is using integration by parts when substitution would be shorter. If one factor is clearly the derivative of an inside function, review U-Substitution Guide before committing to parts.
Calculator check
After solving by hand, use the Integral Calculator to compare steps. If the answer looks different, differentiate both forms. Integration by parts often produces equivalent expressions that simplify in different ways.
Manual practice set
Practice integration by parts with families of related products. Start with \(\int xe^x\,dx\), then try \(\int x^2e^x\,dx\), and finally \(\int x^3e^x\,dx\). The pattern shows why repeated parts works: each differentiation lowers the polynomial degree until the remaining integral is simple.
Next compare \(\int x\sin x\,dx\) and \(\int x\cos x\,dx\). The setup is almost identical, but the antiderivatives of sine and cosine create different signs. Writing \(u\), \(dv\), \(du\), and \(v\) in a four-cell table keeps the signs visible.
How to choose between methods
Before using parts, ask whether substitution would remove the main complication. If one factor is the derivative of an inside expression, substitution is usually cleaner. If one factor becomes simpler when differentiated and the other stays easy to integrate, parts is usually better. This method choice is often more important than the formula itself.
Human review checklist
Check four items: did you choose a \(dv\) that can actually be integrated, did differentiating \(u\) simplify the problem, did you keep the minus sign, and did the remaining integral get easier? After finishing, differentiate your answer. Integration by parts answers often have multiple equivalent forms, so derivative checking is more reliable than matching formatting.
Internal study path
For products where one factor is an inside derivative, read U-Substitution Guide. For rational products, read Partial Fraction Decomposition. For a broader formula review, use The Complete Integration Rules and Formulas Cheat Sheet.
Classroom-style activity
Make a parts table with columns for \(u\), \(dv\), \(du\), \(v\), and the remaining integral. Fill it for \(\int x\ln x\,dx\), \(\int x^2\sin x\,dx\), and \(\int e^x\cos x\,dx\). The last example creates a repeating integral, so it shows that parts sometimes needs algebra after the formula is applied.
Before checking with a calculator, predict whether one round will finish the problem. If \(u\) is a polynomial, repeated differentiation eventually reaches zero. If both functions cycle, such as exponential times sine or cosine, the integral may return to a multiple of itself. Recognizing these patterns makes the method feel organized instead of random.
For editorial review, verify the sign in every parts step. The formula has one subtraction, and later antiderivatives can add another sign. Many polished-looking solutions fail because a negative sign disappeared between lines. A derivative check at the end is the fastest independent test.
Editorial quality gate
This guide is designed to be used with a human check, not as a blind answer source. Before relying on any result, review the notation, the variable, the assumptions, and the final answer type. If a formula contains a bound, a domain restriction, an absolute value, or a convergence condition, that detail should appear in the written reasoning, not only in the final line.
For independent verification, use at least one manual test. Differentiate antiderivatives, substitute endpoints in definite integrals, compare signs with a quick graph estimate, or test a limit by direct substitution before using a special rule. When a calculator result disagrees with handwritten work, first check whether the input was interpreted correctly. Parentheses, variables, and bounds can change the entire problem.
The related calculator link is included for practice and comparison. A productive study workflow is to try the setup on paper, read the guide section that matches the method, run the calculator, and then explain any difference in your own words. That final explanation step is what turns a solved example into durable calculus understanding.
Before you move on
Close the guide by writing one original problem, one method clue, and one verification step. For example, the clue might be an inside derivative, an indeterminate form, a repeated factor, a variable bound, or a physical unit. The verification step might be differentiation, substitution, estimation, graph behavior, or comparison with a known formula. This short habit helps the article become active practice rather than passive reading.
For best results, revisit the examples after a break and solve them without looking at the steps. If the method still feels clear, the guide has done its job.