MHB Integration Formulas: Do We Really Need to Memorize Them?

  • Thread starter Thread starter soroban
  • Start date Start date
  • Tags Tags
    Age
AI Thread Summary
The discussion centers on the preference for simpler integration formulas over more complex methods, particularly regarding integrals of the form ∫du/(u²-a²) and ∫du/(u²+a²). One participant expresses frustration with the modern approach that often requires multiple steps to solve integrals, advocating for the memorization of direct formulas that streamline the process. There is also a mention of the importance of understanding the derivation of these formulas, as well as the implications of domain considerations in calculus. Overall, the conversation highlights a divide between traditional and contemporary teaching methods in calculus integration.
soroban
Messages
191
Reaction score
0

I was taught integration formulas like these:

. . \int\frac{du}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|+C . **

It took up only one brain cell and it saved me
the bother of Partial Fractions (every time).I was taught: .\int\frac{du}{u^2+a^2} \;=\;\tfrac{1}{a}\arctan(\tfrac{u}{a})+C

So that: .\int\frac{dx}{x^2+9} \;=\;\tfrac{1}{3}\arctan(\tfrac{x}{3})+CBut it seems that everyone nowadays is taught:

. . \int\frac{du}{u^2+1} \;=\;\arctan u + C

So that, if you are not given a "1" in there,
you must do some Olympic-level gymnastics.

\int\frac{dx}{x^2+9} \;=\;\int\frac{dx}{9(\frac{x^2}{9}+1)} \;=\;\tfrac{1}{9}\int\frac{dx}{(\frac{x}{3})^2+1}

Let u = \tfrac{x}{3} \quad\Rightarrow\quad dx \:=\:3\,du

Substitute: .\tfrac{1}{9}\int \frac{3\,du}{u^2+1} \;=\;\tfrac{1}{3}\arctan u + C

Back-substitute: .\tfrac{1}{3}\arctan(\tfrac{x}{3})+CAm I the only one who prefers doing one step
. . instead of eight?~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**
By the way, if you've memorized that formula,
. . its sister-formula is easily derived.We have: .\int\frac{du}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|+C

Multiply by -1: .-1\int\frac{du}{u^2-a^2} \;=\;-\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right| + C

. . \int\frac{du}{a^2-u^2} \;=\;\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|^{-1} + C

. . \int\frac{du}{a^2-u^2} \;=\;\frac{1}{2a}\ln\left|\frac{u+a}{u-a}\right| + C

. . \int\frac{du}{a^2-u^2} \;=\;\frac{1}{2a}\ln\left|\frac{a+u}{a-u}\right| + C
 
Physics news on Phys.org
Silly me! And there I was thinking I played smart from deriving all of the above from

$$\int \frac{dx}{x} = \log(x) + C \tag{*}$$

And some complicated log-arctan relationships!

$(*)$ : Actually, I can show that the integral on the left side is the inverse of $\exp(x)$, by a little functional analysis ;)
 
One of the things about "finding a primitive" (i.e., an indefinite integral) is that it really doesn't matter HOW you do it, integration is not a mechanical process.

Personally, I would observe that:

$\dfrac{d}{du}\left(\dfrac{u-a}{u+a}\right) = \dfrac{(u+a) - (u - a)}{(u+a)^2}$

$=\dfrac{2a}{(u+a)^2}$

so the result follows from the chain rule.

Regarding mathbalarka's observation, he makes a good point but *some* posters in this forum feels this is "too much to handle" for beginning calculus students. He ought to say the inverse is more properly:

$\displaystyle \int_0^x \dfrac{1}{t}\ dt$ for $x > 0$, the integral is undefined if we use an interval containing 0.

(Similar concerns result with any integral that evaluates to some inverse function such as $\log$, we have to check the "domain of integration" to ensure our answer is *consistent*...the absolute value signs are not entirely a safeguard against this).
 
"In 2013, after a series of hospitalizations due to magnets, New Zealand government officials permanently banned the sale of those made from neodymium-iron-boron (NIB)." https://www.sciencealert.com/new-zealand-teen-loses-part-of-his-bowel-after-swallowing-nearly-200-magnets "A 13-year-old boy in New Zealand has had part of his bowel surgically removed after he ingested nearly 200 high-powered magnets. (2025)" OK, this teen was a fool, but it is not always a teen, it is not always...
Thread 'RIP Chen Ning Yang (1922-2025)'
https://en.wikipedia.org/wiki/Yang_Chen-Ning ( photo from http://insti.physics.sunysb.edu/~yang/ ) https://www.nytimes.com/2025/10/18/science/chen-ning-yang-dead.html https://www.bbc.com/news/articles/cdxrzzk02plo https://www.cpr.cuhk.edu.hk/en/press/mourning-professor-yang-chen-ning/ https://www.stonybrook.edu/commcms/physics/about/awards_and_prizes/_nobel_and_breakthrough_prizes/_profiles/yangc https://www.stonybrook.edu/commcms/physics/people/_profiles/yangc...
Back
Top