How to Reverse the Product Rule

Let’s bring our attention to one function: $\ln$ .

Theorem 1. $\displaystyle \ln'(x) = 1/x$ .

Proof. Coupling $\ln = \exp^{-1}$ with $\exp' = \exp$ , the inverse function theorem yields

$\displaystyle \ln'(x) = (\exp^{-1})'(x) = \frac{1}{\exp'(\exp^{-1}(x))} = \frac{1}{\exp(\exp^{-1}(x))} = \frac 1x.$

This, in fact, needed to be proven before we derive the useful implications of the reverse chain rule.

How can we calculate $\displaystyle \int \ln(x)\, \mathrm dx$ ?

Theorem 2. $\displaystyle \int \ln(x)\,\mathrm dx = x(\ln(x) - 1) + C$ .

It is trivial to differentiate the right-hand side to get $\ln(x)$ . The real question is how we discovered the right-hand side in the first place.

We need to use integration by parts, which I like to call the reverse product rule.

Theorem 3 (Reverse Product Rule). Let $u\equiv u(x),v\equiv v(x)$ be known differentiable functions. Then

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

where we formally abbreviated $\mathrm du = u'(x)\,\mathrm dx$ and $\mathrm dv = v'(x)\,\mathrm dx$ .

Proof. Reverse the product rule $(uv)' = v u' + u v'$ .

For pedagogical utility, I introduce my students to IS-ID notation:

$\displaystyle \int v'u\, \mathrm dx = \underbrace{v}_{\text{I}}\underbrace{u}_{\text{S}} - \int \underbrace{v}_{\text{I}} \underbrace{u'}_{\text{D}}\, \mathrm dx.$

Just like the reverse chain rule, the utility of this rule does not arise in its theory (which is immediate), but in its pervasive applications.

The crucial trick here is that $u$ , $v$ , and the integral of $vu'$ must be known. Let’s investigate the integral of question:

$\displaystyle \int \ln(x)\,\mathrm dx.$

Here, if we allow $u = 1$ and $v' = \ln(x)$ , then $u' = 0$ , but what is $v$ ? It’s unknown! This choice of $u,v'$ therefore would be useless for this question.

One useful heuristic would be LIATE, which, roughly speaking, summarises the varied difficulty of the knowledge of $v$ . For this blogpost, we will not reference LIATE, but may detail it in the future as an exercise.

Let’s instead work with the choice $v' = 1$ and $u = \ln(x)$ . Integrating $v'$ yields $v = x$ , and differentiating $u$ yields $u' = 1/x$ . Now we know all the required information, and can make (at least partial) progress in computing the integral.

Proof of Theorem 2. Using the reverse product rule (i.e. integrating by parts),

$\displaystyle \begin{aligned} \int 1 \cdot \ln(x)\, \mathrm dx &= \underbrace{x}_{\text{I}}\underbrace{\ln(x)}_{\text{S}} - \int \underbrace{x}_{\text{I}} \underbrace{\frac 1x}_{\text{D}}\, \mathrm dx \\ &= x \ln(x) - \int 1\, \mathrm dx \\ &= x \ln(x) - x + C,\end{aligned}$

where the integration after the first step is trivial.

Alternate Proof of Theorem 2. Alternatively, make the substitution

$u = \ln(x) \quad \Rightarrow \quad \mathrm dx = e^u\, \mathrm du.$

Then by Theorem 3 below,

$\displaystyle \int \ln(x)\, \mathrm dx = \int ue^u\, \mathrm du = (u-1)e^u + C = x(\ln(x) - 1) + C.$

The power of these integration techniques, therefore, is to help us compute integrals that were, up till now, out of reach by our naïve knowledge of reverse-differentiation. The case of integrating $\ln(x)$ can be generalised.

Corollary 1. Suppose $u(x)$ is a known differentiable function such that the anti-derivative of $xu'(x)$ is known. Then

$\displaystyle \int u(x)\, \mathrm dx = xu(x) - \int xu'(x)\, \mathrm dx.$

This next integral explores the situation when both $u$ and $v$ are known, but not necessarily the integral of $vu'$ .

Theorem 3. $\displaystyle \int xe^x\, \mathrm dx = (x-1)e^x + C$ .

Proof. There are two ways to integrate this expression:

$\begin{aligned} \int x \cdot e^x\, \mathrm dx &= \underbrace{\frac{x^2}{2}}_{\text{I}}\underbrace{e^x}_{\text{S}} - \int \underbrace{\frac{x^2}{2}}_{\text{I}} \underbrace{e^x}_{\text{D}}\, \mathrm dx = \frac 12 x^2 e^x - \frac 12 \int x^2 e^x\, \mathrm dx, \\ \int x \cdot e^x\, \mathrm dx &= \underbrace{e^x}_{\text{I}}\underbrace{x}_{\text{S}} - \int \underbrace{e^x}_{\text{I}} \underbrace{1}_{\text{D}}\, \mathrm dx = xe^x - \int e^x\, \mathrm dx. \end{aligned}$

Observe that the first integral is hard to compute, while the second integral gives us the desired answer. The choice of $u,v'$ therefore directly affects the outcome of our answer.

Integrating $\tan^{-1}$ is an interesting problem involving both the reverse product rule and the reverse chain rule.

Theorem 4. $\displaystyle \int \tan^{-1}(x)\, \mathrm dx = x\tan^{-1}(x) - \frac 12 \ln(1+x^2) + C$ .

Proof. Integrating by parts then integrating by substitution,

$\begin{aligned} \int 1 \cdot \tan^{-1}(x)\, \mathrm dx &= \underbrace{x}_{\text{I}}\underbrace{\tan^{-1}(x)}_{\text{S}} - \int \underbrace{x}_{\text{I}} \underbrace{\frac{1}{1 + x^2}}_{\text{D}}\, \mathrm dx \\ &= x \tan^{-1}(x) - \frac 12 \int \frac{2x}{1+x^2}\, \mathrm dx \\ &= x \tan^{-1}(x) - \frac 12 \int \frac{(1+x^2)'}{1+x^2}\, \mathrm dx \\ &= x \tan^{-1}(x) - \frac 12 \ln(1+x^2) + C. \end{aligned}$

Finally, it is entirely possible to apply the reverse product rule twice to obtain a sensible answer.

Theorem 5. For $a, b > 0$ ,

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

Proof. Integrating by parts once (in this special case, the choice of $u,v'$ doesn’t really affect the answer),

$\displaystyle \begin{aligned} \int e^{ax} \cos(bx)\, \mathrm dx &= \underbrace{ \frac{e^{ax}}{a} }_{\text I} \underbrace{ \cos(bx) }_{\text S} - \int \underbrace{ \frac{e^{ax}}{a} }_{\text I} \underbrace{ (-b\sin(bx)) }_{\text D}\, \mathrm dx \\ &= \frac{e^{ax}}{a}\cos(bx) + \frac{b}{a} \int e^{ax} \sin(bx)\, \mathrm dx. \end{aligned}$

Integrating by parts a second time,

$\displaystyle \begin{aligned} \int e^{ax} \sin(bx)\, \mathrm dx &= \underbrace{ \frac{e^{ax}}{a} }_{\text I} \underbrace{ \sin(bx) }_{\text S} - \int \underbrace{ \frac{e^{ax}}{a} }_{\text I} \underbrace{ (b\cos(bx)) }_{\text D}\, \mathrm dx \\ &= \frac{e^{ax}}{a} \sin(bx) - \frac{b}{a} \int e^{ax} \cos(bx)\, \mathrm dx. \end{aligned}$

Back-substituting,

$\displaystyle \begin{aligned} \int e^{ax} \cos(bx)\, \mathrm dx &= \frac{e^{ax}}{a} \cos(bx) + \frac{b}{a} \left( \frac{e^{ax}}{a} \sin(bx) - \frac{b}{a} \int e^{ax} \cos(bx)\, \mathrm dx \right). \end{aligned}$

By algebra,

$\displaystyle \left(1 + \frac{b^2}{a^2}\right) \int e^{ax} \cos(bx)\, \mathrm dx = \frac{ e^{ax} }{a^2} \left( a\cos(bx) + b\sin(bx) \right).$

The result follows by dividing by $(1 + b^2/a^2)$ , then adding the arbitrary constant $+C$ .

And that is how we reverse the product rule to solve even more integration problems—integration by parts, which I call the IS-ID technique.

What good is integration for? How about even defining angles in the first place? What do I mean? Next post.

—Joel Kindiak, 26 Oct 24, 1753H

Responses

Proving the Area of a Circle and More – KindiakMath

October 27, 2024 at 3:06 pm

[…] Integrating by parts and using the chain rule, […]

LikeLike

Calculus Recap (Polytechnic) – KindiakMath

February 13, 2025 at 6:59 pm

[…] Solution. Applying the linearity of integration and using common formulas (here and here), […]

LikeLike

Who Cares about Laplace Transforms? – KindiakMath

March 18, 2025 at 1:34 pm

[…] Proof. Integrating by parts, […]

LikeLike

Several Laborious Integrals – KindiakMath

March 26, 2025 at 11:11 am

[…] Solution. We first evaluate for simplicity. Integrating by parts, […]

LikeLike

Differential Equation Rally – KindiakMath

May 9, 2025 at 1:07 pm

[…] We leave it as an exercise in integration by parts to verify […]

LikeLike

Exponential Fourier Series – KindiakMath

May 14, 2025 at 7:11 pm

[…] Integrating by parts […]

LikeLike

Mid-Term Rally – KindiakMath

June 4, 2025 at 10:09 am

[…] Without Laplace transforms, you will need to integrate by parts to […]

LikeLike

Infinite Addition – KindiakMath

July 27, 2025 at 10:13 pm

[…] Integrating by parts, […]

LikeLike

End-Term Rally – KindiakMath

August 3, 2025 at 10:58 pm

[…] since is odd, we integrate by parts to […]

LikeLike

Kindiak Gauntlet (Differential Equations) – KindiakMath

August 19, 2025 at 4:17 pm

[…] we integrated by parts twice on the right-hand […]

LikeLike

Fourier Orthogonality – KindiakMath

October 6, 2025 at 1:10 pm

[…] for any , two applications of integration by parts […]

LikeLike

Why u-Substitution is So Awesome – KindiakMath

January 5, 2026 at 8:14 pm

[…] Next time, we reverse the product rule. […]

LikeLike

Proving that Pi is Irrational – KindiakMath

January 5, 2026 at 8:34 pm

[…] integrals can be simplified using integration by parts. For […]

LikeLike

Some Integration Drills – KindiakMath

January 5, 2026 at 9:09 pm

[…] Integrating by parts twice, […]

LikeLike

The Classic Secant Integral – KindiakMath

January 5, 2026 at 9:10 pm

[…] Write . Since integrates to , we integrate by parts to […]

LikeLike

Bounded Derivative Integrals – KindiakMath

January 5, 2026 at 9:13 pm

[…] Solution. Integrating by parts, […]

LikeLike

Some Normal Distribution Trivia – KindiakMath

January 5, 2026 at 9:21 pm

[…] , we first integrate by parts to […]

LikeLike

The Wallis Product – KindiakMath

January 5, 2026 at 9:22 pm

[…] to check that and . These observations motivate an expression of in terms of . To that end, we integrate by parts to […]

LikeLike

How to Reverse the Product Rule

Share this:

Responses

Leave a reply to Some Normal Distribution Trivia – KindiakMath Cancel reply