Question

If I compute the integral of (x^2-9)^-1 dx than I get 1/6 * (ln \left| x-3 \right| - ln \left| x+3 \right|) + C. If I compare my answer with wolfram alpha I am kind of confused because the switch my ln \left| x-3 \right| to ln(3-x) - without using the absolute value and the switch the position. Why?

Answer 1

\[\int\limits_{?}^{?}1/(x^2-9)dx = 1/6 (\ln \left| x-3 \right|-\ln \left| x+3 \right|\] (for some reason it is wrong in my initial question). Wolfram Alpha gives the same thing but without the absolute value and a the change the ln (x-3) to ln(3-x). Why?

Answer 2

\[\int \frac{1}{(x^2-9)} dx =\int \frac{1}{6}\frac{1}{(x-3)}-\frac{1}{6}\frac{1}{(x+3)} dx\]\[= \frac{1}{6}ln{|x-3|}-\frac{1}{6}ln{|x+3|} +C\]\[= \frac{1}{6}ln{\frac{|x-3|}{|x+3|}}+C\]

I get the same answer as you :)

Answer 3

So why does Wolfram Alpha gives an answer without the absolute value and the switched position from ln(x-3) to ln(3-x)? I always thought Wolfram knows mathematics better than me :D

Answer 4

I think it has to do with substitution Wolfram used.

Answer 5

But it does not look like an alternative answer - more like a wrong answer. In the answers we get we can plug in any value for x but in wolframs answer you have to look out that the ln always get´s feed with a number greater than 0.