integrate ln function

**twilightstr** · February 7th 2009, 01:46 PM

integrate ln(x^2-1)

**galactus** · February 7th 2009, 02:32 PM

Use the rules of exponents to rewrite as

$ln(x^{2}-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)$

**mr fantastic** · February 7th 2009, 02:36 PM

Originally Posted by galactus

Use the rules of exponents to rewrite as

$ln(x^{2}-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)$

But care needs to be taken with domain issues since the original function is defined for x > 1 or x < -1 whereas the re-write has problems when x < -1 ....

**galactus** · February 7th 2009, 03:29 PM

Yep. I was being lazy and careless.

**twilightstr** · February 7th 2009, 10:49 PM

so how should I go about solving this problem?

**HallsofIvy** · February 8th 2009, 07:11 AM

Do exactly what galactus said but be careful like mr fantastic said!

$ln(x^2- 1)$ 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.

**javax** · February 8th 2009, 08:33 AM

Or you can go with integration by parts

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

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

so we have

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

correct me anyone if I did anything wrong.

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