integrate ln(x^2-1)

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- Feb 7th 2009, 01:46 PMtwilightstrintegrate ln function
integrate ln(x^2-1)

- Feb 7th 2009, 02:32 PMgalactus
Use the rules of exponents to rewrite as

$\displaystyle ln(x^{2}-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)$ - Feb 7th 2009, 02:36 PMmr fantastic
- Feb 7th 2009, 03:29 PMgalactus
Yep. I was being lazy and careless.

- Feb 7th 2009, 10:49 PMtwilightstr
so how should I go about solving this problem?

- Feb 8th 2009, 07:11 AMHallsofIvy
Do exactly what galactus said but be careful like mr fantastic said!

$\displaystyle ln(x^2- 1)$ is an even function so you really only need to integrate for x> 1, where galactus' method has no problem, and then use the same integral for x< -1. - Feb 8th 2009, 08:33 AMjavax
Or you can go with integration by parts

$\displaystyle \int{\ln(x^2-1)}$

sub. $\displaystyle \ln(x^2-1)=u \Rightarrow \frac{2x dx}{x^2-1}=du; dv = dx \Rightarrow v = x;$

so we have

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

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

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

correct me anyone if I did anything wrong.