Calculus I · Unit 3A · lesson

Integration by Parts

Learn Integration by Parts through clear explanation, worked examples, visual reasoning, checks, and connected integral-calculus practice.

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Section overview

Computing integrals

What this section is building

Learn Integration by Parts through clear explanation, worked examples, visual reasoning, checks, and connected integral-calculus practice.

Notice

Substitution reverses a chain rule, parts reverses a product rule, and algebraic or trigonometric rewrites expose a recognizable derivative pattern.

Decide

Simplify first; look for an inner derivative; then consider parts, identities, trigonometric substitution, or partial fractions in a deliberate order.

Avoid

Choosing a method by surface appearance, transforming only part of the differential, or accepting a more complicated integral than the one you started with.

Concept

Learning objectives

Derive the formula from the product rule, choose $u$ and $dv$ , and use the method on products and disguised products.

Integration by Parts

Explanation

Integration by parts redistributes a product

Integration by parts comes from integrating the product rule. It is useful when an integrand is a product whose factors become easier in different directions: one factor simplifies when differentiated, while the other can be integrated. The formula $\int u\,dv=uv-\int v\,du$ does not eliminate work automatically; it trades the original integral for a hopefully simpler one.

Choosing $u$ and $dv$ is therefore the central judgment. Logarithmic and inverse-trigonometric factors are usually chosen as $u$ because they are difficult to integrate directly but manageable to differentiate. Polynomial factors often improve when differentiated. After applying the formula, compare the new integral with the old one. If it is more complicated, reconsider the choice rather than marching deeper into algebraic wilderness out of loyalty to the first idea.

From the product rule,

(uv)'=u'v+uv'.

Integrating and rearranging gives

\boxed{\int u\,dv=uv-\int v\,du}.

Guided walkthrough

Polynomial times exponential

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

Answer reveal

Worked solution

Your setup or answer

Write a real attempt before opening the supplied answer.

Worked example

A disguised product

To integrate $\ln x$ , write $\ln x\cdot1$ . Let $u=\ln x$ and $dv=dx$ . Then

\int\ln x\,dx=x\ln x-\int1\,dx=x\ln x-x+C.

Interactive checku3a-parts-01

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

Your work stays on this device. No account or AI grader is used.

Show hint

Let $u=x$ and $dv=e^x dx$ .

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

**Visual study**
Integration by parts from the product rule. Show product rule rearranged and integrated, with u and dv roles.

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

Integration by Parts

Explanation

Integration by parts redistributes a product

From the product rule,

(uv)'=u'v+uv'.

Integrating and rearranging gives

\boxed{\int u\,dv=uv-\int v\,du}.

Guided walkthrough

Polynomial times exponential

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

Answer reveal

Worked solution

Your setup or answer

Write a real attempt before opening the supplied answer.

Worked example

A disguised product

To integrate $\ln x$ , write $\ln x\cdot1$ . Let $u=\ln x$ and $dv=dx$ . Then

\int\ln x\,dx=x\ln x-\int1\,dx=x\ln x-x+C.

Interactive checku3a-parts-01

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

Your work stays on this device. No account or AI grader is used.

Show hint

Let $u=x$ and $dv=e^x dx$ .

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Read this graph as text

Integration by parts from the product rule. The product rule is rearranged to produce integral u dv = uv - integral v du. Show product rule rearranged and integrated, with u and dv roles. Do not present LIATE as a theorem; it is a heuristic.

Every relationship in integration by parts from the product rule uses written labels together with distinct line styles, markers, or fill patterns; color is never the only carrier of meaning.

Why it matters: Show product rule rearranged and integrated, with u and dv roles.

Visual study

Integration by parts from the product rule. Show product rule rearranged and integrated, with u and dv roles.

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

Integration by Parts

Related resources

Original instruction with traceable references.

Integration by Parts

Related resources

Original instruction with traceable references.