Prove that f(x)=ln(x + (x^2+1)^0.5 ) is an odd function...solved it by rationalizing but need another way if possible pls :/

Here's an alternative proof:

Prove that f(x)=ln(x + (x^2+1)^0.5 ) is an odd function... :/

If this is an odd function then

 f(-x)=-f(x)

and since x^2= (-x)^2

that means you are suppose to prove that 

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

Now 

$ln(x+\sqrt{x^2+1})$ $=sinh^{-1}(x)$

is the inverse hyperbolic sine function so it is definitely odd.  But I do not know how to prove this.

I shall flag other mathematicians to take a look for you.  You should get an answer but it meight take a day or so. ://

$-\ln(x+\sqrt{x^2+1}) = \ln\left(\dfrac{1}{x+\sqrt{x^2+1}}\right) =$

$\ln\left(\dfrac{1}{x+\sqrt{x^2+1}} \dfrac{x-\sqrt{x^2+1}}{x-\sqrt{x^2+1}}\right) = \ln\left( \dfrac{x-\sqrt{x^2+1}}{x^2-(x^2+1)}\right)=$

$\ln\left(-(x-\sqrt{x^2+1})\right) =\ln\left(-x + \sqrt{x^2+1}\right)$

$\mbox{so }f(-x)=-f(x) \mbox{ and f is odd.}$

Thanks Rom,

I should of thought of that I guess.    

I suppose things often look easy when someone shows you how to do it.    

I think the questioner knew how to solve it Rom's way (see the question title).  He/she was looking for a possible alternative method!

Thank you guys alot...

Thank you Alan for your alternative answer, it helped me alot!   

I'm new to this site so excuse my late reply