The Central Idea
The product rule says
\[ \frac{d}{dx}[u(x)v(x)] = u(x)v'(x)+v(x)u'(x). \]In differential form,
\[ d(uv)=u\,dv+v\,du. \]Rearranging gives
\[ u\,dv=d(uv)-v\,du. \]Integrating both sides produces the integration-by-parts formula:
\[ \boxed{ \int u\,dv = uv-\int v\,du }. \]Integration by Parts Reverses the Product Rule
Substitution reverses the chain rule.
Integration by parts reverses the product rule.
The structural comparison is
\[ \boxed{ \text{product rule} \longleftrightarrow \text{integration by parts} }. \]Instead of differentiating a product, we use a known product-rule identity to rewrite an integral of a product.
The Four Pieces
In
\[ \int u\,dv = uv-\int v\,du, \]we choose:
- \(u\), the factor we will differentiate;
- \(dv\), the factor we will integrate.
Then compute:
\[ du=u'(x)\,dx \]and
\[ v=\int dv. \]The goal is to make
\[ \int v\,du \]simpler than the original integral.
Example: \(\int x e^x\,dx\)
Consider
\[ \int x e^x\,dx. \]Ordinary substitution does not simplify this product.
Choose
\[ u=x \]and
\[ dv=e^x\,dx. \]Then
\[ du=dx \]and
\[ v=e^x. \]Completing the First Example
Apply the formula:
\[ \int x e^x\,dx = x e^x-\int e^x\,dx. \]Therefore,
\[ \boxed{ \int x e^x\,dx = x e^x-e^x+C }. \]Factoring gives
\[ \boxed{ e^x(x-1)+C }. \]Checking by Differentiation
Differentiate
\[ e^x(x-1). \]By the product rule,
\[ e^x(x-1)+e^x. \]Simplify:
\[ xe^x. \]The original integrand returns.
Why the Choice of \(u\) Matters
Integration by parts is not merely a formula to substitute into.
We must choose \(u\) and \(dv\) strategically.
A useful choice makes
\[ du \]simpler than \(u\), while making
\[ v=\int dv \]easy to find.
The new integral
\[ \int v\,du \]should be simpler than the one we started with.
The LIATE Guideline
A common heuristic for choosing \(u\) is LIATE:
- Logarithmic functions
- Inverse trigonometric functions
- Algebraic functions
- Trigonometric functions
- Exponential functions
When two different types of factors appear, choose \(u\) from the category that occurs earlier in the list when that choice simplifies the problem.
LIATE Is a Guideline, Not a Law
LIATE helps with many common problems, but it is not a theorem.
The real criterion is:
\[ \boxed{ \text{choose }u\text{ so that }\int v\,du \text{ becomes easier} }. \]If a different choice clearly simplifies the problem more effectively, use that choice.
Example: \(\int x\sin x\,dx\)
Choose
\[ u=x \]and
\[ dv=\sin x\,dx. \]Then
\[ du=dx \]and
\[ v=-\cos x. \]Apply integration by parts:
\[ \int x\sin x\,dx = -x\cos x+\int\cos x\,dx. \]Therefore,
\[ \boxed{ -x\cos x+\sin x+C }. \]Example: \(\int x\cos x\,dx\)
Choose
\[ u=x, \qquad dv=\cos x\,dx. \]Then
\[ du=dx, \qquad v=\sin x. \]So
\[ \int x\cos x\,dx = x\sin x-\int\sin x\,dx. \]Therefore,
\[ \boxed{ x\sin x+\cos x+C }. \]Integration by Parts When No Product Is Visible
Consider
\[ \int \ln x\,dx. \]Write the integrand as
\[ \ln x\cdot1. \]Choose
\[ u=\ln x, \qquad dv=dx. \]Then
\[ du=\frac1x\,dx, \qquad v=x. \]Completing \(\int\ln x\,dx\)
Apply the formula:
\[ \int\ln x\,dx = x\ln x-\int x\frac1x\,dx. \]Simplify:
\[ x\ln x-\int1\,dx. \]Therefore,
\[ \boxed{ x\ln x-x+C }. \]Example: \(\int\arctan x\,dx\)
Write
\[ \arctan x = (\arctan x)(1). \]Choose
\[ u=\arctan x, \qquad dv=dx. \]Then
\[ du=\frac{1}{1+x^2}\,dx, \qquad v=x. \]Therefore,
\[ \int\arctan x\,dx = x\arctan x - \int\frac{x}{1+x^2}\,dx. \]Combining Methods
The remaining integral
\[ \int\frac{x}{1+x^2}\,dx \]is handled by substitution:
\[ u=1+x^2. \]This gives
\[ \frac12\ln(1+x^2). \]Therefore,
\[ \boxed{ \int\arctan x\,dx = x\arctan x - \frac12\ln(1+x^2)+C }. \]Integration techniques often work together.
Example: \(x\ln x\)
Evaluate
\[ \int x\ln x\,dx. \]LIATE suggests choosing
\[ u=\ln x \]and
\[ dv=x\,dx. \]Then
\[ du=\frac1x\,dx \]and
\[ v=\frac{x^2}{2}. \]Completing \(\int x\ln x\,dx\)
Apply the formula:
\[ \int x\ln x\,dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\frac1x\,dx. \]Simplify:
\[ \frac{x^2}{2}\ln x - \frac12\int x\,dx. \]Therefore,
\[ \boxed{ \frac{x^2}{2}\ln x - \frac{x^2}{4} +C }. \]Repeated Integration by Parts
Some products require integration by parts more than once.
Consider
\[ \int x^2e^x\,dx. \]Choose
\[ u=x^2, \qquad dv=e^x\,dx. \]Then
\[ du=2x\,dx, \qquad v=e^x. \]So
\[ \int x^2e^x\,dx = x^2e^x - 2\int xe^x\,dx. \]Applying the Method Again
From the earlier example,
\[ \int xe^x\,dx = xe^x-e^x. \]Therefore,
\[ \int x^2e^x\,dx = x^2e^x - 2(xe^x-e^x). \]Simplify:
\[ \boxed{ e^x(x^2-2x+2)+C }. \]Why Polynomial–Exponential Products Simplify
Each differentiation reduces the degree of a polynomial:
\[ x^3\to3x^2\to6x\to6\to0. \]Meanwhile, \(e^x\) remains easy to integrate:
\[ e^x\to e^x. \]This makes repeated integration by parts especially effective for products of polynomials and exponential functions.
Polynomial–Trigonometric Products
The same strategy works for expressions such as
\[ x^2\sin x \]or
\[ x^3\cos x. \]Differentiate the polynomial until it disappears, while repeatedly integrating the trigonometric factor.
These problems often require several applications of integration by parts.
Tabular Integration by Parts
When integration by parts must be repeated several times, a tabular method can organize the work efficiently.
The idea is to:
- differentiate one factor repeatedly;
- integrate the other factor repeatedly;
- alternate signs;
- multiply diagonally.
This is most useful when repeated differentiation eventually reaches zero.
Example: Tabular Setup for \(x^3e^x\)
| Sign | Differentiate \(x^3\) | Integrate \(e^x\) |
|---|---|---|
| \(+\) | \(x^3\) | \(e^x\) |
| \(-\) | \(3x^2\) | \(e^x\) |
| \(+\) | \(6x\) | \(e^x\) |
| \(-\) | \(6\) | \(e^x\) |
| \(0\) | \(e^x\) |
Reading the diagonal products gives
\[ \boxed{ \int x^3e^x\,dx = e^x(x^3-3x^2+6x-6)+C }. \]When the Tabular Method Is Appropriate
The tabular method is particularly efficient when one factor is a polynomial and the other is easy to integrate repeatedly.
It is less useful when repeated differentiation does not simplify the chosen \(u\), or when the integrations in the other column become more difficult.
The table is an organizational shortcut, not a separate theorem.
Cyclic Integration by Parts
Some integrals return to themselves after repeated integration by parts.
This may look like a dead end, but it creates an algebraic equation that can be solved for the original integral.
A classic example is
\[ \int e^x\cos x\,dx. \]Example: First Integration by Parts
Let
\[ I=\int e^x\cos x\,dx. \]Choose
\[ u=\cos x, \qquad dv=e^x\,dx. \]Then
\[ du=-\sin x\,dx, \qquad v=e^x. \]Therefore,
\[ I=e^x\cos x+\int e^x\sin x\,dx. \]Apply Integration by Parts Again
Let
\[ J=\int e^x\sin x\,dx. \]Choose
\[ u=\sin x, \qquad dv=e^x\,dx. \]Then
\[ du=\cos x\,dx, \qquad v=e^x. \]So
\[ J=e^x\sin x-\int e^x\cos x\,dx. \]Thus,
\[ J=e^x\sin x-I. \]Solve for the Original Integral
Substitute
\[ J=e^x\sin x-I \]into
\[ I=e^x\cos x+J. \]Then
\[ I=e^x\cos x+e^x\sin x-I. \]Therefore,
\[ 2I=e^x(\sin x+\cos x). \]Hence,
\[ \boxed{ \int e^x\cos x\,dx = \frac{e^x}{2}(\sin x+\cos x)+C }. \]Integration by Parts for Definite Integrals
The definite-integral form is
\[ \boxed{ \int_a^b u\,dv = \left[uv\right]_a^b - \int_a^b v\,du }. \]The product term \(uv\) must be evaluated at both endpoints.
Example: Definite Integration by Parts
Evaluate
\[ \int_0^1 xe^x\,dx. \]Use
\[ u=x, \qquad dv=e^x\,dx. \]Then
\[ du=dx, \qquad v=e^x. \]Therefore,
\[ \int_0^1xe^x\,dx = [xe^x]_0^1 - \int_0^1e^x\,dx. \]Completing the Definite Example
Evaluate the product term:
\[ [xe^x]_0^1=e. \]Evaluate the remaining integral:
\[ \int_0^1e^x\,dx=e-1. \]Therefore,
\[ e-(e-1)=1. \]Hence,
\[ \boxed{1}. \]Example: \(\int_1^e\ln x\,dx\)
Since
\[ \int\ln x\,dx=x\ln x-x, \]we have
\[ \int_1^e\ln x\,dx = [x\ln x-x]_1^e. \]At \(x=e\),
\[ e(1)-e=0. \]At \(x=1\),
\[ 1(0)-1=-1. \]Therefore,
\[ \boxed{1}. \]Substitution or Integration by Parts?
A useful first question is whether the integrand contains a clear inner-function-and-derivative pattern.
If yes, substitution is often the better method.
If the integrand is a product where differentiating one factor simplifies it and integrating the other is straightforward, integration by parts may be better.
Compare:
\[ \int 2xe^{x^2}\,dx \]which suggests substitution, with
\[ \int xe^x\,dx, \]which suggests integration by parts.
Methods Can Work Together
Some problems require more than one integration method.
For example,
\[ \int \arctan x\,dx \]begins with integration by parts, then uses substitution on the remaining rational integral.
Integration should be viewed as a toolbox rather than as a collection of isolated procedures.
Not Every Product Requires Integration by Parts
A visible product does not automatically mean integration by parts is the right method.
For example,
\[ \int 2x(x^2+1)^4\,dx \]is a product, but substitution is much more direct.
Look for structure before selecting a method.
Choosing \(dv\)
The factor assigned to \(dv\) must be easy to integrate.
Common choices include
\[ e^x\,dx, \qquad \sin x\,dx, \qquad \cos x\,dx, \qquad x^n\,dx. \]Avoid choosing \(dv\) so that finding \(v\) is harder than the original problem.
Choosing \(u\)
A good \(u\) typically becomes simpler when differentiated.
Examples:
\[ x^3\to3x^2, \] \[ \ln x\to\frac1x, \] \[ \arctan x\to\frac1{1+x^2}. \]The derivative does not always need to be shorter in appearance, but it should improve the new integral.
Example of an Unhelpful Choice
In
\[ \int xe^x\,dx, \]suppose we choose
\[ u=e^x \]and
\[ dv=x\,dx. \]Then
\[ du=e^x\,dx, \qquad v=\frac{x^2}{2}. \]The new integral becomes
\[ \frac12\int x^2e^x\,dx, \]which is harder than the original.
Track the Minus Sign Carefully
The formula is
\[ \boxed{ \int u\,dv=uv-\int v\,du }. \]If \(v\) itself is negative, the signs may combine.
For example, when
\[ dv=\sin x\,dx, \]we get
\[ v=-\cos x. \]Writing each stage before simplifying reduces sign errors.
Pull Out Constant Factors First
If
\[ \int 5x e^x\,dx \]appears, first write
\[ 5\int xe^x\,dx. \]Then apply integration by parts to the essential product.
This keeps the bookkeeping cleaner.
Simplify Algebraically Before Choosing a Method
Consider
\[ \int \frac{x^2+x}{x}\,dx. \]Simplify first:
\[ \frac{x^2+x}{x}=x+1. \]Then
\[ \boxed{ \int(x+1)\,dx = \frac{x^2}{2}+x+C }. \]Integration by parts would be unnecessary.
Common Error: Forgetting the Minus Sign
The correct formula is
\[ \boxed{ \int u\,dv = uv-\int v\,du }. \]Writing
\[ uv+\int v\,du \]does not reverse the product rule correctly.
Common Error: Mixing Up \(du\) and \(dv\)
Once \(u\) and \(dv\) are chosen:
- differentiate \(u\) to obtain \(du\);
- integrate \(dv\) to obtain \(v\).
Do not differentiate both pieces or integrate both pieces.
Common Error: Creating a Harder Integral
A valid choice of \(u\) and \(dv\) can still be strategically poor.
After one application, inspect
\[ \int v\,du. \]If it is clearly more complicated, reconsider the original choice.
Common Error: Stopping Too Early
In repeated integration-by-parts problems, the first application may not finish the integral.
For
\[ \int x^2e^x\,dx, \]the first step produces another integral requiring integration by parts:
\[ \int xe^x\,dx. \]Continue until the remaining integral is elementary.
Common Error: Treating a Repeated Integral as Failure
If the original integral reappears after repeated integration by parts, do not automatically restart.
Move the repeated integral to the other side and solve algebraically.
Cyclic integrals are designed to close this way.
Common Error: Forgetting to Evaluate \(uv\) at the Bounds
For definite integrals,
\[ \int_a^b u\,dv = [uv]_a^b-\int_a^b v\,du. \]The boundary term is
\[ \boxed{ u(b)v(b)-u(a)v(a) }. \]Common Error: Adding \(+C\) to a Definite Integral
As always:
- indefinite integrals end with \(+C\);
- definite integrals evaluate to numbers and do not.
A Reliable Procedure for Integration by Parts
- Simplify the integrand algebraically when useful.
- Check whether substitution is more natural first.
- Choose \(u\) and \(dv\).
- Differentiate \(u\) to find \(du\).
- Integrate \(dv\) to find \(v\).
- Apply \(\int u\,dv=uv-\int v\,du\).
- Simplify the new integral.
- Repeat integration by parts if necessary.
- Use substitution or another method if the remaining integral requires it.
- Add \(+C\) for indefinite integrals.
- Differentiate the final result to verify it.
A Reliable Procedure for Definite Integrals
- Choose \(u\) and \(dv\).
- Compute \(du\) and \(v\).
- Write \([uv]_a^b-\int_a^b v\,du\).
- Evaluate the boundary term carefully.
- Evaluate the remaining definite integral.
- Combine the results.
- Do not add \(+C\).
- Check the sign and approximate magnitude when possible.
A Fast Method-Selection Checklist
Consider integration by parts when:
- the integrand is a product;
- one factor simplifies when differentiated;
- the other factor is easy to integrate;
- a logarithm or inverse trigonometric function appears alone;
- repeated differentiation of a polynomial will eventually reach zero;
- substitution does not reveal a simpler reverse-chain-rule pattern.
Example: Full Workflow with \(x^2\cos x\)
Evaluate
\[ \int x^2\cos x\,dx. \]Choose
\[ u=x^2, \qquad dv=\cos x\,dx. \]Then
\[ du=2x\,dx, \qquad v=\sin x. \]Therefore,
\[ \int x^2\cos x\,dx = x^2\sin x-2\int x\sin x\,dx. \]Completing the Workflow
We already know
\[ \int x\sin x\,dx = -x\cos x+\sin x. \]Substitute:
\[ x^2\sin x - 2(-x\cos x+\sin x). \]Therefore,
\[ \boxed{ x^2\sin x+2x\cos x-2\sin x+C }. \]Example: Full Definite Workflow
Evaluate
\[ \int_0^{\pi/2}x\cos x\,dx. \]Choose
\[ u=x, \qquad dv=\cos x\,dx. \]Then
\[ du=dx, \qquad v=\sin x. \]Thus,
\[ \int_0^{\pi/2}x\cos x\,dx = [x\sin x]_0^{\pi/2} - \int_0^{\pi/2}\sin x\,dx. \]Completing the Definite Workflow
The boundary term is
\[ \frac{\pi}{2}(1)-0=\frac{\pi}{2}. \]Also,
\[ \int_0^{\pi/2}\sin x\,dx=1. \]Therefore,
\[ \boxed{ \frac{\pi}{2}-1 }. \]Practice 1: Polynomial–Exponential
Evaluate:
\[ \int xe^x\,dx. \]Practice 2: Polynomial–Trigonometric
Evaluate:
\[ \int x\cos x\,dx. \]Practice 3: Logarithm
Evaluate:
\[ \int\ln x\,dx. \]Practice 4: Algebraic–Logarithmic
Evaluate:
\[ \int x\ln x\,dx. \]Practice 5: Repeated Integration by Parts
Evaluate:
\[ \int x^2e^x\,dx. \]Practice 6: Repeated Trigonometric Product
Evaluate:
\[ \int x^2\sin x\,dx. \]Practice 7: Inverse Trigonometric Function
Evaluate:
\[ \int\arctan x\,dx. \]Practice 8: Definite Integral
Evaluate:
\[ \int_0^1 xe^x\,dx. \]Practice 9: Definite Logarithmic Integral
Evaluate:
\[ \int_1^e\ln x\,dx. \]Practice 10: Choose \(u\)
For
\[ \int x^3e^x\,dx, \]which factor is the natural choice for \(u\)?
Practice 11: Choose the Method
Which method is more natural for
\[ \int 2xe^{x^2}\,dx: \]substitution or integration by parts?
Practice 12: Identify the Poor Choice
In
\[ \int xe^x\,dx, \]explain why choosing
\[ u=e^x, \qquad dv=x\,dx \]is strategically poor.
Practice 13: Identify the Error
A student writes
\[ \int u\,dv = uv+\int v\,du. \]Correct the formula.
Practice 14: Boundary Term
In a definite integration-by-parts problem on \([a,b]\), how is the term
\[ uv \]handled?
Practice Answers
Practice 1
Choose
\[ u=x, \qquad dv=e^x\,dx. \]Then
\[ du=dx, \qquad v=e^x. \]Therefore,
\[ \boxed{ xe^x-e^x+C }. \]Practice 2
Choose
\[ u=x, \qquad dv=\cos x\,dx. \]Then
\[ du=dx, \qquad v=\sin x. \]Thus,
\[ \boxed{ x\sin x+\cos x+C }. \]Practice 3
Write
\[ \ln x=(\ln x)(1). \]Choose
\[ u=\ln x, \qquad dv=dx. \]Therefore,
\[ \boxed{ x\ln x-x+C }. \]Practice 4
Choose
\[ u=\ln x, \qquad dv=x\,dx. \]Then
\[ du=\frac1x\,dx, \qquad v=\frac{x^2}{2}. \]Therefore,
\[ \boxed{ \frac{x^2}{2}\ln x-\frac{x^2}{4}+C }. \]Practice 5
Applying integration by parts twice gives
\[ \boxed{ e^x(x^2-2x+2)+C }. \]Practice 6
First,
\[ \int x^2\sin x\,dx = -x^2\cos x+2\int x\cos x\,dx. \]Since
\[ \int x\cos x\,dx = x\sin x+\cos x, \]we obtain
\[ \boxed{ -x^2\cos x+2x\sin x+2\cos x+C }. \]Practice 7
Choose
\[ u=\arctan x, \qquad dv=dx. \]The result is
\[ \boxed{ x\arctan x-\frac12\ln(1+x^2)+C }. \]Practice 8
Using integration by parts,
\[ \int_0^1xe^x\,dx = [xe^x]_0^1-[e^x]_0^1. \]Therefore,
\[ \boxed{1}. \]Practice 9
Since
\[ \int\ln x\,dx=x\ln x-x, \]we have
\[ [x\ln x-x]_1^e. \]Therefore,
\[ \boxed{1}. \]Practice 10
Choose the algebraic factor:
\[ \boxed{u=x^3}. \]Repeated differentiation will reduce it to zero.
Practice 11
Substitution is more natural because
\[ u=x^2 \]gives
\[ du=2x\,dx. \]Therefore,
\[ \boxed{\text{substitution}}. \]Practice 12
With
\[ u=e^x, \qquad dv=x\,dx, \]the new integral becomes proportional to
\[ \int x^2e^x\,dx, \]which is harder than the original.
The better choice is
\[ \boxed{ u=x,\qquad dv=e^x\,dx }. \]Practice 13
The correct formula is
\[ \boxed{ \int u\,dv = uv-\int v\,du }. \]Practice 14
Evaluate the product at the endpoints:
\[ \boxed{ [uv]_a^b = u(b)v(b)-u(a)v(a) }. \]What to Remember
Integration by parts reverses the product rule:
\[ \boxed{ \int u\,dv = uv-\int v\,du }. \]The strategic goal is to choose \(u\) and \(dv\) so that the new integral
\[ \int v\,du \]is easier than the original.
LIATE is a useful guide for choosing \(u\), especially when logarithmic, inverse trigonometric, algebraic, trigonometric, and exponential factors appear together.
Repeated integration by parts is especially effective when repeated differentiation simplifies one factor until it disappears.
The most important habit is:
do not apply the formula mechanically—choose the split so that the problem becomes simpler.
Where We Go Next
Integration by parts handles many products by reversing the product rule. The next lesson, Integration by Partial Fractions, develops a different strategy for rational functions.
Instead of reversing a differentiation rule, partial fractions decomposes a complicated rational expression into simpler fractions whose antiderivatives are already familiar.