Welcome to our community

Be a part of something great, join today!

Am I from the Stone Age?

soroban

Well-known member
Feb 2, 2012
409

I was taught integration formulas like these:

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

It took up only one brain cell and it saved me
the bother of Partial Fractions (every time).


I was taught: .[tex]\int\frac{du}{u^2+a^2} \;=\;\tfrac{1}{a}\arctan(\tfrac{u}{a})+C[/tex]

So that: .[tex]\int\frac{dx}{x^2+9} \;=\;\tfrac{1}{3}\arctan(\tfrac{x}{3})+C[/tex]


But it seems that everyone nowadays is taught:

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

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

[tex]\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}[/tex]

Let [tex]u = \tfrac{x}{3} \quad\Rightarrow\quad dx \:=\:3\,du[/tex]

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

Back-substitute: .[tex]\tfrac{1}{3}\arctan(\tfrac{x}{3})+C[/tex]


Am 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: .[tex]\int\frac{du}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|+C[/tex]

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

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

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

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

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
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 ;)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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).