🎯⭐ INTERACTIVE LESSON

Integration by Parts

Learn step-by-step with interactive practice!

Integration by Parts - Complete Interactive Lesson

Part 1: The Formula

∫ Integration by Parts

Part 1 of 7 — The Formula & LIATE Rule

Integration by parts is the integration counterpart of the product rule for derivatives. It’s essential for AP Calculus BC and appears on virtually every exam.

Part	Topic
1	The Formula & LIATE Rule
2	Tabular (Column) Method
3	Cycling (Boomerang) Problems
4	Definite Integrals with IBP
5	Special Cases — Inverse Trig & Logarithms
6	Problem-Solving Workshop
7	Comprehensive Review & Assessment

The Integration by Parts Formula

Starting from the product rule: $\frac{d}{dx}[u \cdot v] = u\frac{dv}{dx} + v\frac{du}{dx}$

Integrating both sides and rearranging:

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

Think of it as “swapping” one integral for a (hopefully) simpler one.

Key Fact: Integration by parts is the go-to technique when the integrand is a product of two different types of functions (e.g., polynomial × exponential).

The LIATE Rule for Choosing $u$

The hardest part is deciding which factor to call $u$ and which to call $dv$ . Use the LIATE priority:

Priority	Type	Examples	Why choose as $u$ ?
1st	Logarithmic	$\ln x$ , $\log_2 x$	Differentiates to algebraic
2nd	Inverse trig	$\arctan x$ , $\arcsin x$	Differentiates to algebraic
3rd	Algebraic	$x^2$ , $3x+1$	Differentiates to simpler polynomial
4th	Trigonometric	$\sin x$ , $\cos x$	Stays trig but doesn’t grow
5th	Exponential	$e^x$ , $2^x$	Integrates easily; stays the same type

AP Tip: LIATE works for ~95% of IBP problems. The idea: $u$ should get simpler when differentiated, while $dv$ should be easy to integrate.

Worked Example — $\int x e^x\,dx$

Step	Action	Result
1	Choose $u$ and $dv$	$u = x$ (A), $dv = e^x\,dx$ (E)
2	Differentiate $u$	$du = dx$
3	Integrate $dv$	$v = e^x$
4	Apply formula	$xe^x - \int e^x\,dx$
5	Evaluate remaining integral	$xe^x - e^x + C$

$\boxed{\int x e^x\,dx = e^x(x - 1) + C}$

Verification: $\frac{d}{dx}[e^x(x-1)] = e^x(x-1) + e^x = xe^x$ ✔

Applying the Formula

LIATE Selection Practice

Compute an IBP Integral

Key Takeaways — Part 1

Concept	Details
IBP Formula	$\int u\,dv = uv - \int v\,du$
LIATE Rule	Log > Inverse trig > Algebraic > Trig > Exponential
Goal	Transform a hard integral into an easier one
Key integrals	$\int xe^x\,dx = e^x(x-1) + C$

Coming Up: Part 2 introduces the Tabular Method — a shortcut for polynomial × exponential/trig integrals that eliminates repetitive IBP steps.

Part 2: Tabular Method

∫ Integration by Parts

Part 2 of 7 — Tabular (Column) Method

The tabular method is a shortcut for integrating products of the form (polynomial) × (easy-to-integrate function). Instead of repeatedly applying IBP, you organize everything in a table.

How the Tabular Method Works

Place the polynomial in the “Differentiate” column (it eventually reaches 0)
Place the other factor in the “Integrate” column
Alternate signs: $+, -, +, -, \ldots$
Multiply diagonally and sum

Example: $\int x^2 e^x\,dx$

Sign	Differentiate	Integrate
$+$	$x^2$	$e^x$
$-$	$2x$	$e^x$
$+$	$2$	$e^x$
$-$	$0$	$e^x$

Reading diagonally: $+x^2 e^x - 2xe^x + 2e^x + C$

$\boxed{\int x^2 e^x\,dx = e^x(x^2 - 2x + 2) + C}$

Key Fact: The tabular method works whenever one factor differentiates to zero (polynomials). It saves enormous time on the AP exam.

Tabular Method with Trig

Example: $\int x^3 \cos x\,dx$

Sign	Differentiate	Integrate
$+$	$x^3$	$\cos x$
$-$	$3x^2$	$\sin x$
$+$	$6x$	$-\cos x$
$-$	$6$	$-\sin x$
$+$	$0$	$\cos x$

$\int x^3\cos x\,dx = x^3\sin x + 3x^2\cos x - 6x\sin x - 6\cos x + C$

When Does Tabular NOT Work?

Works ✔	Doesn’t work ✘
$\int x^n e^{ax}\,dx$	$\int e^x \sin x\,dx$ (neither goes to 0)
$\int x^n \sin(ax)\,dx$	$\int \ln x \cdot \arctan x\,dx$
$\int x^n \cos(ax)\,dx$	$\int \frac{\ln x}{x^2}\,dx$ (use standard IBP)

Tabular Method Practice

Identify the Method

Tabular Computation

Key Takeaways — Part 2

Concept	Details
Tabular method	Organize IBP in columns: Sign, Differentiate, Integrate
When to use	Polynomial × exponential or polynomial × trig
Sign pattern	$+, -, +, -, \ldots$ alternating
Reading	Multiply diagonally across columns, then sum

Coming Up: Part 3 covers cycling problems — what happens when IBP brings the original integral back (the boomerang technique).

Part 3: Cycling (Boomerang) Problems

∫ Integration by Parts

Part 3 of 7 — Cycling (Boomerang) Problems

Some IBP integrals don’t simplify — instead, after two applications the original integral reappears. When this happens, you solve algebraically for the unknown integral.

The Boomerang Technique

When does cycling occur? When both factors regenerate under repeated differentiation and integration:

$e^{ax} \cdot \sin(bx)$ or $e^{ax} \cdot \cos(bx)$

Full Worked Example: $\int e^x \sin x\,dx$

Let $I = \int e^x \sin x\,dx$

IBP #1: $u = \sin x$ , $dv = e^x\,dx$ $I = e^x \sin x - \int e^x \cos x\,dx$

IBP #2: $u = \cos x$ , $dv = e^x\,dx$ $I = e^x \sin x - \left[e^x \cos x - \int e^x(-\sin x)\,dx\right]$ $I = e^x \sin x - e^x \cos x - \int e^x \sin x\,dx$ $I = e^x \sin x - e^x \cos x - I$

Solve for $I$ : $2I = e^x(\sin x - \cos x)$

$\boxed{\int e^x \sin x\,dx = \frac{e^x(\sin x - \cos x)}{2} + C}$

Key Fact: You MUST use the same type of choice for $u$ in both applications (both times pick trig, or both times pick exponential). Mixing causes an infinite loop.

General Formula for $e^{ax}$ × Trig

$\boxed{\int e^{ax}\sin(bx)\,dx = \frac{e^{ax}(a\sin(bx) - b\cos(bx))}{a^2 + b^2} + C}$

$\boxed{\int e^{ax}\cos(bx)\,dx = \frac{e^{ax}(a\cos(bx) + b\sin(bx))}{a^2 + b^2} + C}$

Integral	$a$	$b$	Denominator $a^2+b^2$
$\int e^x \sin x\,dx$	1	1	2
$\int e^{2x} \cos(3x)\,dx$	2	3	13
$\int e^{-x} \sin(2x)\,dx$	$-1$	2	5

AP Tip: Memorizing the general formula can save 3–4 minutes on a free-response question. But you should know how to derive it via IBP cycling.

Cycling IBP Practice

Identify the Technique

Cycling Computation

Key Takeaways — Part 3

Concept	Details
Cycling occurs	When both factors regenerate ( $e^{ax}$ × trig)
Strategy	Apply IBP twice, then solve for $I$ algebraically
General sine formula	$\frac{e^{ax}(a\sin bx - b\cos bx)}{a^2+b^2} + C$
General cosine formula	$\frac{e^{ax}(a\cos bx + b\sin bx)}{a^2+b^2} + C$
Critical rule	Use the SAME $u$ -type both times

Coming Up: Part 4 applies IBP to definite integrals — including evaluating bounds correctly.

Part 4: Definite Integrals with IBP

∫ Integration by Parts

Part 4 of 7 — Definite Integrals with IBP

On the AP exam, many IBP problems involve definite integrals. You can either find the antiderivative first, then evaluate at the bounds, or carry the bounds through the entire process.

Definite Integral IBP Formula

$\boxed{\int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du}$

Strategy Options

Approach	When to Use
Find antiderivative, then plug in bounds	Simpler integrals; cleaner algebra
Carry bounds through every step	Avoids needing the general antiderivative

AP Tip: On free-response questions, show each step clearly. Write the $[uv]_a^b$ term explicitly before evaluating.

Worked Example 1: $\int_0^1 xe^x\,dx$

Step	Work
$u = x$ , $dv = e^x\,dx$	$du = dx$ , $v = e^x$
Apply formula	$[xe^x]_0^1 - \int_0^1 e^x\,dx$
Evaluate $[uv]$	$(1 \cdot e) - (0 \cdot 1) = e$
Remaining integral	$[e^x]_0^1 = e - 1$
Final answer	$e - (e-1) = 1$

$\boxed{\int_0^1 xe^x\,dx = 1}$

Worked Example 2: $\int_1^e (\ln x)^2\,dx$

Step	Work
$u = (\ln x)^2$ , $dv = dx$	$du = \frac{2\ln x}{x}\,dx$ , $v = x$
Apply formula	$[x(\ln x)^2]_1^e - \int_1^e 2\ln x\,dx$
Evaluate $[uv]$	$e(1)^2 - 1(0)^2 = e$
Second IBP on $\int \ln x\,dx$	$x\ln x - x + C$
Evaluate	$e - 2[x\ln x - x]_1^e = e - 2[(e-e)-(0-1)]$
Simplify	$e - 2(1) = e - 2$

$\boxed{\int_1^e (\ln x)^2\,dx = e - 2}$

Definite IBP Practice

Step-by-Step Evaluation

Exact Computation

Key Takeaways — Part 4

Integral	Value
$\int_0^1 xe^x\,dx$	$1$
$\int_1^e \ln x\,dx$	$1$
$\int_0^{\pi} x\sin x\,dx$	$\pi$
$\int_1^e (\ln x)^2\,dx$	$e - 2$

Key Fact: Many definite IBP integrals yield surprisingly clean answers. Always simplify fully before reporting your answer.

Coming Up: Part 5 covers special cases including inverse trig and logarithmic IBP integrals.

Part 5: Special Cases

∫ Integration by Parts

Part 5 of 7 — Special Cases: Inverse Trig & Logarithms

Some functions don’t have obvious antiderivatives, but they DO have known derivatives. For these, we set the tricky function as $u$ and let $dv = dx$ .

The “ $dv = dx$ ” Strategy

When the integrand has no obvious product structure, set:

$u =$ the function (so you can differentiate it)
$dv = dx$ (so $v = x$ )

This works beautifully for:

Function	$u$	$du$
$\ln x$	$\ln x$	$\frac{1}{x}\,dx$
$\arctan x$	$\arctan x$	$\frac{1}{1+x^2}\,dx$
$\arcsin x$	$\arcsin x$	$\frac{1}{\sqrt{1-x^2}}\,dx$
$(\ln x)^2$	$(\ln x)^2$	$\frac{2\ln x}{x}\,dx$

Key Fact: L and I in LIATE always become $u$ . Their derivatives produce algebraic expressions that pair nicely with $v = x$ .

Essential Results

1. $\int \ln x\,dx$

$u = \ln x$ , $dv = dx \Rightarrow du = \frac{1}{x}\,dx$ , $v = x$

$\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - x + C$

$\boxed{\int \ln x\,dx = x(\ln x - 1) + C}$

2. $\int \arctan x\,dx$

$u = \arctan x$ , $dv = dx \Rightarrow du = \frac{1}{1+x^2}\,dx$ , $v = x$

$\int \arctan x\,dx = x\arctan x - \int \frac{x}{1+x^2}\,dx$

The remaining integral: let $w = 1+x^2$ , $dw = 2x\,dx$ :

$\boxed{\int \arctan x\,dx = x\arctan x - \frac{1}{2}\ln(1+x^2) + C}$

3. $\int \arcsin x\,dx$

$u = \arcsin x$ , $dv = dx \Rightarrow du = \frac{1}{\sqrt{1-x^2}}\,dx$ , $v = x$

$\int \arcsin x\,dx = x\arcsin x - \int \frac{x}{\sqrt{1-x^2}}\,dx$

Let $w = 1 - x^2$ : remaining integral $= \sqrt{1-x^2}$

$\boxed{\int \arcsin x\,dx = x\arcsin x + \sqrt{1-x^2} + C}$

Higher Powers of $\ln x$

$\int (\ln x)^2\,dx$

$u = (\ln x)^2$ , $dv = dx$

$= x(\ln x)^2 - 2\int \ln x\,dx = x(\ln x)^2 - 2(x\ln x - x) + C$

$\boxed{\int (\ln x)^2\,dx = x[(\ln x)^2 - 2\ln x + 2] + C}$

Reduction formula (for reference): $\int (\ln x)^n\,dx = x(\ln x)^n - n\int (\ln x)^{n-1}\,dx$

Special Cases Practice

Identify the Setup

Numerical Evaluation

Key Takeaways — Part 5

Integral	Result
$\int \ln x\,dx$	$x(\ln x - 1) + C$
$\int \arctan x\,dx$	$x\arctan x - \frac{1}{2}\ln(1+x^2) + C$
$\int \arcsin x\,dx$	$x\arcsin x + \sqrt{1-x^2} + C$
$\int (\ln x)^2\,dx$	$x[(\ln x)^2 - 2\ln x + 2] + C$

AP Tip: These results are worth memorizing — they appear frequently in free-response questions and save substantial time.

Coming Up: Part 6 is a problem-solving workshop with mixed IBP challenges.

Part 6: Practice Workshop

∫ Integration by Parts

Part 6 of 7 — Problem-Solving Workshop

This part is a mixed-practice workshop. Every problem requires identifying the correct IBP approach, then executing it. Think before you compute!

Decision Flowchart

Integrand Type	Method
Polynomial × $e^{ax}$ or trig	Tabular method
$e^{ax}$ × trig	Cycling (boomerang)
$\ln x$ , $\arctan x$ , $\arcsin x$ alone	$dv = dx$ strategy
Polynomial × $\ln x$	Standard IBP ( $u = \ln x$ )
Anything else with a product	Standard IBP with LIATE

Mixed IBP Practice — Round 1

Mixed IBP Practice — Round 2

Method Identification

Definite Integral Challenge

Key Takeaways — Part 6

Problem Type	Method	Time on AP Exam
Poly × exp/trig	Tabular	~2 min
$e^{ax}$ × trig	Cycling	~3 min
Inverse trig or log alone	$dv = dx$	~2 min
Mixed product	LIATE + standard	~3 min

AP Tip: If you get stuck, try a different $u$ / $dv$ split. There’s often more than one path to the answer.

Coming Up: Part 7 is the comprehensive review and assessment covering all IBP techniques.

Part 7: Final Assessment

∫ Integration by Parts — Review

Part 7 of 7 — Comprehensive Review & Assessment

This final part tests your mastery of all IBP techniques: the basic formula, LIATE, tabular method, cycling, definite integrals, and special cases.

Complete Reference Table

Integral	Antiderivative
$\int xe^x\,dx$	$e^x(x-1) + C$
$\int x\cos x\,dx$	$x\sin x + \cos x + C$
$\int x\sin x\,dx$	$-x\cos x + \sin x + C$
$\int x^2 e^x\,dx$	$e^x(x^2 - 2x + 2) + C$
$\int e^x\sin x\,dx$	$\frac{e^x(\sin x - \cos x)}{2} + C$
$\int e^x\cos x\,dx$	$\frac{e^x(\sin x + \cos x)}{2} + C$
$\int \ln x\,dx$	$x(\ln x - 1) + C$
$\int \arctan x\,dx$	$x\arctan x - \frac{1}{2}\ln(1+x^2) + C$
$\int \arcsin x\,dx$	$x\arcsin x + \sqrt{1-x^2} + C$

Assessment — Conceptual

Assessment — Computational

Method Selection Review

Final Calculation

Integration by Parts — Complete! ✅

You’ve mastered:

✔ The IBP formula and LIATE rule
✔ Tabular method for polynomial × exp/trig
✔ Cycling technique for $e^{ax}$ × trig
✔ Definite integrals with IBP
✔ Special cases (inverse trig, logs)
✔ Mixed problem identification

AP Exam Frequency

IBP Type	Likelihood on AP BC Exam
Basic IBP or tabular	Almost certain (MC + FRQ)
Cycling ( $e^{ax}$ × trig)	Common in MC
Special cases ( $\ln$ , inverse trig)	Frequent in FRQ
Definite IBP	Very common

Key Fact: Integration by parts appears on every AP Calculus BC exam. Master all five approaches and you’ll handle any IBP problem confidently.

← Back to Standard Lesson

Integration by Parts

Integration by Parts - Complete Interactive Lesson

Part 1: The Formula

∫ Integration by Parts

The Integration by Parts Formula

The LIATE Rule for Choosing uuu

Worked Example — ∫xex dx\int x e^x\,dx∫xexdx

Key Takeaways — Part 1

Part 2: Tabular Method

∫ Integration by Parts

How the Tabular Method Works

Example: ∫x2ex dx\int x^2 e^x\,dx∫x2exdx

Tabular Method with Trig

When Does Tabular NOT Work?

Key Takeaways — Part 2

Part 3: Cycling (Boomerang) Problems

∫ Integration by Parts

The Boomerang Technique

Full Worked Example: ∫exsin⁡x dx\int e^x \sin x\,dx∫exsinxdx

General Formula for eaxe^{ax}eax × Trig

Key Takeaways — Part 3

Part 4: Definite Integrals with IBP

∫ Integration by Parts

Definite Integral IBP Formula

Strategy Options

Worked Example 1: ∫01xex dx\int_0^1 xe^x\,dx∫01​xexdx

Worked Example 2: ∫1e(ln⁡x)2 dx\int_1^e (\ln x)^2\,dx∫1e​(lnx)2dx

Key Takeaways — Part 4

Part 5: Special Cases

∫ Integration by Parts

The “dv=dxdv = dxdv=dx” Strategy

Essential Results

Higher Powers of ln⁡x\ln xlnx

Key Takeaways — Part 5

Part 6: Practice Workshop

∫ Integration by Parts

Decision Flowchart

Key Takeaways — Part 6

Part 7: Final Assessment

∫ Integration by Parts — Review

Complete Reference Table

Integration by Parts — Complete! ✅

AP Exam Frequency

Integration by Parts

Integration by Parts - Complete Interactive Lesson

Part 1: The Formula

∫ Integration by Parts

The Integration by Parts Formula

The LIATE Rule for Choosing uuu

Worked Example — ∫xex dx\int x e^x\,dx∫xexdx

Key Takeaways — Part 1

Part 2: Tabular Method

∫ Integration by Parts

How the Tabular Method Works

Example: ∫x2ex dx\int x^2 e^x\,dx∫x2exdx

Tabular Method with Trig

When Does Tabular NOT Work?

Key Takeaways — Part 2

Part 3: Cycling (Boomerang) Problems

∫ Integration by Parts

The Boomerang Technique

Full Worked Example: ∫exsin⁡x dx\int e^x \sin x\,dx∫exsinxdx

General Formula for eaxe^{ax}eax × Trig

Key Takeaways — Part 3

Part 4: Definite Integrals with IBP

∫ Integration by Parts

Definite Integral IBP Formula

Strategy Options

Worked Example 1: ∫01xex dx\int_0^1 xe^x\,dx∫01​xexdx

Worked Example 2: ∫1e(ln⁡x)2 dx\int_1^e (\ln x)^2\,dx∫1e​(lnx)2dx

Key Takeaways — Part 4

Part 5: Special Cases

∫ Integration by Parts

The “dv=dxdv = dxdv=dx” Strategy

Essential Results

Higher Powers of ln⁡x\ln xlnx

Key Takeaways — Part 5

Part 6: Practice Workshop

∫ Integration by Parts

Decision Flowchart

The LIATE Rule for Choosing $u$

Worked Example — $\int x e^x\,dx$

Example: $\int x^2 e^x\,dx$

Full Worked Example: $\int e^x \sin x\,dx$

General Formula for $e^{ax}$ × Trig

Worked Example 1: $\int_0^1 xe^x\,dx$

Worked Example 2: $\int_1^e (\ln x)^2\,dx$

The “ $dv = dx$ ” Strategy

Higher Powers of $\ln x$

The LIATE Rule for Choosing $u$

Worked Example — $\int x e^x\,dx$

Example: $\int x^2 e^x\,dx$

Full Worked Example: $\int e^x \sin x\,dx$

General Formula for $e^{ax}$ × Trig

Worked Example 1: $\int_0^1 xe^x\,dx$

Worked Example 2: $\int_1^e (\ln x)^2\,dx$

The “ $dv = dx$ ” Strategy

Higher Powers of $\ln x$