Integration by Parts: Choosing the First and Second Functions

Integration by parts becomes much less confusing when you stop asking only, “What comes first in LIATE?”

A better question is:

Which choice will make the next integral easier than the original one?

That is the real heart of the method.

LIATE is useful. It gives you a sensible starting point. But integration by parts is not meant to be a blind alphabetical exercise. It is a method based on judgment. You choose one part to differentiate and one part to integrate, then check whether the new integral has become simpler.

Let us build that idea carefully with examples, decisions, and a few traps that students often miss.

The Basic Formula

Integration by parts comes from the product rule of differentiation.

The formula is:

Integral of u dv = uv - Integral of v du

In school notation, you may also see it written like this:

Integral of first function times second function dx
= first function times integral of second function dx
- Integral of [differentiation of first function times integral of second function dx] dx

Both forms mean the same thing.

The practical job is to split the given product into two roles:

Role	What happens to it
First function, or u	Differentiate it
Second function, or dv part	Integrate it

So the choice matters.

If the first function becomes simpler after differentiation, the question usually opens up. If the second function is easy to integrate, the formula remains manageable.

What LIATE Actually Means

LIATE is a memory order used to choose the first function.

Letter	Function type	Examples
L	Logarithmic	log x, ln x
I	Inverse trigonometric	tan^-1 x, sin^-1 x
A	Algebraic	x, x^2, 3x + 1
T	Trigonometric	sin x, cos x, tan x
E	Exponential	e^x, a^x

The function that comes earlier in LIATE is usually chosen as the first function.

For example, in:

Integral of x e^x dx

x is algebraic and e^x is exponential.

Algebraic comes before exponential in LIATE, so choose:

First function: x
Second function: e^x

That works because:

d/dx (x) = 1

The first function becomes simpler immediately.

The Two Checks Before You Apply the Formula

Before writing the full solution, pause for two quick checks.

Check	Question to ask	Why it matters
Differentiation check	Will the first function become simpler?	This reduces the difficulty of the next integral
Integration check	Can I integrate the second function easily?	This keeps the formula usable

If both answers are yes, your choice is probably good.

If one answer is no, think again.

For example:

Integral of x cos x dx

Try:

First function: x
Second function: cos x

Now check:

d/dx (x) = 1
Integral of cos x dx = sin x

Both are simple. This is a good choice.

But if you choose:

First function: cos x
Second function: x

then:

d/dx (cos x) = -sin x
Integral of x dx = x^2 / 2

The new integral becomes:

Integral of (x^2 / 2)(-sin x) dx

That is harder than the original. The choice has moved the question in the wrong direction.

Example 1: Algebraic With Exponential

Evaluate:

Integral of x e^x dx

Using LIATE, choose:

u = x
dv = e^x dx

Then:

du = dx
v = e^x

Apply the formula:

Integral of x e^x dx = x e^x - Integral of e^x dx

So:

Integral of x e^x dx = x e^x - e^x + C

Factor if you want a neater final answer:

Integral of x e^x dx = e^x(x - 1) + C

Why did this choice work?

Because x became 1, while e^x stayed easy to integrate.

Example 2: Algebraic With Trigonometric

Evaluate:

Integral of x sin x dx

Using LIATE:

u = x
dv = sin x dx

Then:

du = dx
v = -cos x

Apply the formula:

Integral of x sin x dx = x(-cos x) - Integral of (-cos x)(dx)

So:

Integral of x sin x dx = -x cos x + Integral of cos x dx

Therefore:

Integral of x sin x dx = -x cos x + sin x + C

Notice the sign carefully. Since the integral of sin x is -cos x, the first term becomes negative.

Example 3: Logarithmic Function With Algebraic Function

Evaluate:

Integral of x log x dx

Here log x is logarithmic and x is algebraic.

Logarithmic comes before algebraic in LIATE, so choose:

u = log x
dv = x dx

Then:

du = 1/x dx
v = x^2 / 2

Apply the formula:

Integral of x log x dx
= (log x)(x^2 / 2) - Integral of (x^2 / 2)(1/x) dx

Simplify the remaining integral:

Integral of x log x dx
= (x^2 log x) / 2 - Integral of x / 2 dx

So:

Integral of x log x dx
= (x^2 log x) / 2 - x^2 / 4 + C

This example shows why logarithmic functions are usually chosen first. They become simpler when differentiated, while they are not always pleasant to integrate directly.

Example 4: When There Is Only One Visible Function

Sometimes integration by parts is needed even when the question does not look like a product.

For example:

Integral of log x dx

There is only log x visible. Still, you can treat the other factor as 1.

Write:

Integral of log x dx = Integral of log x . 1 dx

Choose:

u = log x
dv = 1 dx

Then:

du = 1/x dx
v = x

Apply the formula:

Integral of log x dx = x log x - Integral of x(1/x) dx

So:

Integral of log x dx = x log x - Integral of 1 dx

Therefore:

Integral of log x dx = x log x - x + C

Example 5: Inverse Trigonometric Function

Evaluate:

Integral of tan^-1 x dx

Write it as:

Integral of tan^-1 x . 1 dx

Choose:

u = tan^-1 x
dv = 1 dx

Then:

du = 1 / (1 + x^2) dx
v = x

Apply the formula:

Integral of tan^-1 x dx
= x tan^-1 x - Integral of x / (1 + x^2) dx

Now solve the remaining integral.

For:

Integral of x / (1 + x^2) dx

the derivative of 1 + x^2 is 2x, so:

Integral of x / (1 + x^2) dx = (1/2) log(1 + x^2)

Therefore:

Integral of tan^-1 x dx
= x tan^-1 x - (1/2) log(1 + x^2) + C

This is a good example of integration by parts creating a smaller integral that can then be finished by substitution.

When You May Need Integration by Parts Twice

If the algebraic function has a higher power, one round may not finish the question.

For example:

Integral of x^2 e^x dx

Choose:

u = x^2
dv = e^x dx

Then:

du = 2x dx
v = e^x

Apply the formula:

Integral of x^2 e^x dx = x^2 e^x - Integral of 2x e^x dx

The remaining integral:

Integral of 2x e^x dx

still needs integration by parts.

This is normal. The method is working because the power has dropped from x^2 to x.

After the second round, the power will drop again and the question will finish.

A Quick Pattern for Repeated Parts

For products like:

x^2 e^x
x^3 sin x
x^4 cos x

choose the algebraic part as the first function.

Why?

Because repeated differentiation will eventually reduce it to zero.

First function	After one differentiation	After two differentiations	What happens
x	1	0	Finishes quickly
x^2	2x	2	Needs more rounds
x^3	3x^2	6x	Needs still more rounds

This is why the tabular method, if your teacher has shown it, works nicely for repeated integration by parts with algebraic functions.

But even without the tabular method, the thinking is the same:

Pick the part that becomes simpler again and again.

When LIATE Can Mislead You

LIATE is not wrong, but it is incomplete.

It does not think for you.

Consider:

Integral of x^3 e^(x^2) dx

If you only look at x^3 and e^(x^2), you might try a simple integration by parts choice. But the better observation is:

x^3 e^(x^2) = x^2 . x e^(x^2)

Now choose:

u = x^2
dv = x e^(x^2) dx

Why is this smarter?

Because:

Integral of x e^(x^2) dx

is easy by substitution.

So the second function is not just e^(x^2). It is x e^(x^2), because that part can be integrated neatly.

This is why you should not follow LIATE mechanically. First look for an inner derivative, a simple substitution, or a factor that should stay with the second function.

How to Decide in Exam-Style Questions

Use this order of thinking:

1. Is the question a product of two different types of functions?
2. If yes, does LIATE suggest a first function?
3. Will differentiating that first function make it simpler?
4. Can the remaining second function be integrated easily?
5. After applying the formula, is the new integral easier?

If the answer to the last question is no, your choice may need revision.

Here is a compact decision table.

Integral type	Usually choose first	Reason
x e^x, x^2 e^x	Algebraic part	Power reduces on differentiation
x sin x, x cos x	Algebraic part	Power reduces and trig integrates easily
log x alone	log x	Treat the second part as 1 dx
x log x	log x	Log becomes 1/x
tan^-1 x alone	tan^-1 x	Treat the second part as 1 dx
e^x sin x	Either one, but be consistent	The integral may return to itself

The Special Case of e^x sin x or e^x cos x

Some integrals do not become simpler in the usual one-way manner.

For example:

Integral of e^x sin x dx

If you apply integration by parts once, you get another integral involving e^x cos x.

If you apply it again, the original integral comes back.

That may feel strange at first, but it is useful. When the original integral returns, you collect like terms and solve for it.

The lesson is:

Not every integration by parts question ends because a power becomes zero.
Some end because the original integral repeats and can be moved to one side.

So if you see a product of exponential and trigonometric functions, expect two rounds and a final algebraic adjustment.

The Most Common Mistakes

Here are the mistakes that cost marks again and again.

Mistake	What goes wrong	Better habit
Choosing u only from memory	The remaining integral may become harder	Preview the next integral
Forgetting dv includes dx	The formula becomes messy	Write dv = ... dx clearly
Integrating the first function instead of differentiating it	Roles get reversed	First function is differentiated
Losing negative signs	Trig answers become wrong	Write v carefully before substituting
Stopping after one round too early	Higher powers remain unfinished	Repeat if the new integral is still parts-friendly
Ignoring substitution	A useful factor may be separated wrongly	Look for inner derivatives before deciding

A Simple Practice Set

Try deciding only the first and second functions before solving.

Question	First function	Second function
Integral of x e^x dx	x	e^x
Integral of x cos x dx	x	cos x
Integral of x^2 log x dx	log x	x^2
Integral of tan^-1 x dx	tan^-1 x	1
Integral of x^2 sin x dx	x^2	sin x

Now add the reason beside each answer.

That reason is more important than the table itself. It trains your mind to choose, not guess.

The Final Test

Before committing to a choice, ask:

After this choice, is the remaining integral simpler, familiar, or repeatable?

If it is simpler, continue.

If it is familiar, solve it using substitution or a standard integral.

If it repeats the original integral, collect terms and solve.

If it becomes more difficult, go back and choose differently.

That is the difference between using LIATE and depending on LIATE.

Frequently Asked Questions

What is the first function in integration by parts?

The first function is the part you choose to differentiate. In Integral of u dv, it is u. A good first function usually becomes simpler after differentiation.

What is the second function in integration by parts?

The second function is the part you integrate. In Integral of u dv, it is included in dv. It should be something whose integral you can find easily.

Is LIATE always correct?

No. LIATE is a useful guide, but it is not a fixed law. After using it, you should still check whether the remaining integral becomes easier.

What should I choose as the first function in Integral of x e^x dx?

Choose x as the first function and e^x as the second function. Differentiating x gives 1, and integrating e^x gives e^x, so the next integral becomes easy.

How do I use integration by parts when there is only one function?

Write the function as multiplied by 1. For example, write Integral of log x dx as Integral of log x . 1 dx. Then choose log x as the first function and 1 dx as dv.

Why do some questions need integration by parts twice?

Some first functions, such as x^2 or x^3, do not become fully simple after one differentiation. Each round reduces the power, so you may need to repeat the method.

What should I do if the new integral becomes harder?

Stop and change your choice. The purpose of integration by parts is to create an easier integral, not just to apply a formula.

Can substitution and integration by parts be used in the same question?

Yes. Many questions use integration by parts first and then substitution for the remaining integral. You should always look for a familiar inner derivative after applying the formula.