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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]