## Friday, August 17, 2007

### Another 'LN' post. (i.e. log_e!)

I unfortunately seem to not get on too well, with the pleasant folks who live in the town called Logarithms. I guess what I'm seeking, from those who do get on with those pesky folks(!), is confirmation.

The question was:

$\displaystyle \int \frac{2+x}{(1+x)^2} dx$

So, I rewrote it as:

$\displaystyle \int \frac{1+1+x}{(1+x)^2} dx= \int \frac{1+x}{(1+x)^2}dx + \int \frac{1}{(1+x)^2}dx$

Integrating this gave me,

$\displaystyle \int \frac{2+x}{(1+x)^2} dx= \frac{1}{2}\ln (1+x)^2 -\frac{1}{1+x} +c$

So what's the problem you ask? Well it is not exactly a problem, but a niggling doubt. Is it correct to simplify that answer in the following way?

$\displaystyle \frac{1}{2}\ln (1+x)^2 = \ln(1+x)$

I'm thinking that the answer is yes, since $\ln x^2= 2\ln x$, but why do I always have a problem with this!! I mean, when I differentiate 2x, I do so without thinking twice. But these logs annoy me. :( They seem to have been haunting me since college as well, I mean why does the stinking power have to come down for? The graphs of ln(x^2) and 2ln(x) seems to be different as well!

(So does the 1/2 cancel?) :o

#### 1 comment:

beans said...

I wonder - how did these log rules come about into business! I've had the answer confirmed, but I still don't trust these logs!

Might as well post the way it can be checked on another post - can't use LaTeX here you see.