After learning the product rule in differentiation, a natural question arises:

If differentiation has a product rule, does integration have a reverse process?

The answer is yes.

That reverse process is called integration by parts.

At first, the method may appear to be another formula that must be memorized. However, just as substitution is the reverse of the chain rule, integration by parts is simply the product rule running backward.

Once this connection becomes clear, the technique becomes much easier to understand.

Revisiting the product rule

Suppose we have two functions:

$u(x)$

and

$v(x)$

The product rule states:

$\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$

Or, more compactly:

$\boxed{(uv)'=uv'+vu'}$

This formula tells us how to differentiate the product of two functions.

Turning the product rule around

Rewrite the product rule:

$d(uv)=u,dv+v,du$

Now rearrange it:

$u,dv=d(uv)-v,du$

Integrate both sides:

$\int u,dv=\int d(uv)-\int v,du$

The first integral is straightforward:

$\int d(uv)=uv$

Therefore:

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

This is the integration-by-parts formula.

What does the formula mean?

Integration by parts doesn’t magically solve difficult integrals.

Instead, it transforms one integral into another.

We exchange a complicated integral for one that is hopefully easier.

The strategy is simple:

Separate the integrand into two parts.
Choose one part to become $u$ .
Choose the remaining part to become $dv$ .
Differentiate $u$ .
Integrate $dv$ .
Apply the formula.

Why would we ever do this?

Consider:

$\int xe^x,dx$

Neither the power rule nor substitution seems helpful.

The problem contains two different kinds of functions:

A polynomial:

$x$

An exponential function:

$e^x$

Integration by parts allows us to simplify the product.

Example 1: Integrating $xe^x$

Choose:

$u=x$

Then:

$du=dx$

Choose:

$dv=e^x,dx$

Integrate:

$v=e^x$

Apply the formula:

$\int xe^x,dx=xe^x-\int e^x,dx$

Evaluate the remaining integral:

$=xe^x-e^x+C$

Factor:

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

How do we choose $u$ ?

This is one of the biggest challenges.

A useful guideline is the LIATE rule.

Choose $u$ according to the following priority:

Priority	Function type
L	Logarithmic
I	Inverse trigonometric
A	Algebraic
T	Trigonometric
E	Exponential

Functions near the top usually become $u$ .

Functions near the bottom usually become $dv$ .

Example 2: Integrating $x\sin x$

Choose:

$u=x$

Then:

$du=dx$

Choose:

$dv=\sin x,dx$

Integrate:

$v=-\cos x$

Apply the formula:

$\int x\sin x,dx=-x\cos x+\int\cos x,dx$

Evaluate:

$\boxed{\int x\sin x,dx=-x\cos x+\sin x+C}$

Example 3: Integrating $\ln x$

How do we integrate a logarithm?

Rewrite the integral:

$\int\ln x,dx=\int1\cdot\ln x,dx$

Choose:

$u=\ln x$

Then:

$du=\frac1x,dx$

Choose:

$dv=dx$

Integrate:

$v=x$

Apply the formula:

$\int\ln x,dx=x\ln x-\int1,dx$

Therefore:

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

Without integration by parts, this integral would be difficult to evaluate.

A geometric interpretation

Think of two people carrying a heavy object.

One person represents:

$u$

The other represents:

$dv$

Instead of carrying the entire load simultaneously, one person transfers part of the load to the other.

The work is redistributed.

Integration by parts does exactly the same thing.

It redistributes mathematical complexity.

A business example

Suppose a company’s advertising expenditure is:

$A(t)=t$

and customer engagement grows exponentially:

$E(t)=e^t$

Total accumulated impact can be modeled by:

$\int te^t,dt$

Applying integration by parts:

$=e^t(t-1)+C$

The technique allows us to analyze interactions between two different growth processes.

A physics example

Suppose force changes according to:

$F(t)=t\sin t$

Total work can be calculated using:

$\int t\sin t,dt$

Again, integration by parts provides a solution.

Physics often involves products of multiple changing quantities.

This makes integration by parts an essential tool.

Integration by substitution versus integration by parts

Technique	Reverse of
Substitution	Chain rule
Integration by parts	Product rule

Substitution simplifies nested functions.

Integration by parts simplifies products.

A simple checklist

Whenever you encounter a difficult integral:

Ask Question 1:

Is one function contained inside another?

If yes, try substitution.

Ask Question 2:

Are two functions multiplied together?

If yes, try integration by parts.

The deeper philosophical idea

Differentiation breaks mathematical objects into smaller pieces.

Integration reconstructs them.

Substitution reconstructs composite functions.

Integration by parts reconstructs products.

Both techniques reveal the beautiful symmetry hidden within calculus.

The formula you should remember

Instead of memorizing:

$\int u,dv=uv-\int v,du$

remember the product rule:

$\frac{d}{dx}(uv)=uv'+vu'$

The integration formula emerges naturally.

Conclusion

Integration by parts is not an isolated technique.

It is the product rule running backward.

Whenever two functions are multiplied together, integration by parts allows us to transfer complexity from one function to another.

Perhaps the simplest way to summarize the idea is this:

The product rule takes a product apart.

Integration by parts puts the product back together.

Understanding this relationship transforms integration by parts from a memorized formula into a powerful and intuitive mathematical tool.

Integration by Parts: Reversing the Product Rule

Revisiting the product rule

Turning the product rule around

What does the formula mean?

Why would we ever do this?

Example 1: Integrating $xe^x$

How do we choose $u$ ?

Example 2: Integrating $x\sin x$

Example 3: Integrating $\ln x$

A geometric interpretation

A business example

A physics example

Integration by substitution versus integration by parts

A simple checklist

The deeper philosophical idea

The formula you should remember

Conclusion

Like this:

DavidsonNext: AP® Calculus: Challenging Concepts from Calculus AB & Calculus BC

Calculus 1A: Differentiation

Calculus 1B: Integration

Calculus 1C: Coordinate Systems & Infinite Series

Revisiting the product rule

Turning the product rule around

What does the formula mean?

Why would we ever do this?

Example 1: Integrating

How do we choose ?

Example 2: Integrating

Example 3: Integrating

A geometric interpretation

A business example

A physics example

Integration by substitution versus integration by parts

A simple checklist

The deeper philosophical idea

The formula you should remember

Conclusion

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Like this:

DavidsonNext: AP® Calculus: Challenging Concepts from Calculus AB & Calculus BC

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Calculus 1A: Differentiation

Calculus 1B: Integration

Calculus 1C: Coordinate Systems & Infinite Series

Example 1: Integrating $xe^x$

How do we choose $u$ ?

Example 2: Integrating $x\sin x$

Example 3: Integrating $\ln x$