Okay first post here not sure how to do this...but I've been stuck on a two probelms...I just scanned them and am posting them as a picture...
first one is..
They both are asking to be simplified...
1)
2)
Thank's for the help!
start by canceling the 2's, then ...
$\displaystyle \frac{1}{x + \sqrt{x^2+1}} \left( 1 + \frac{x}{\sqrt{x^2+1}}\right)$
$\displaystyle \frac{1}{x + \sqrt{x^2+1}} \left( \frac{\sqrt{x^2+1}}{\sqrt{x^2+1}} + \frac{x}{\sqrt{x^2+1}}\right)
$
$\displaystyle \frac{1}{x + \sqrt{x^2+1}} \left(\frac{x + \sqrt{x^2+1}}{\sqrt{x^2+1}}\right)
$
can you finish?
multiply the single terms by the (-1)'s ...
$\displaystyle \frac{(e^x + e^{-x})^2 - (e^x - e^{-x})^2}{(e^x + e^{-x})^2}$
$\displaystyle \frac{[(e^x + e^{-x}) - (e^x - e^{-x})] \cdot [(e^x + e^{-x}) + (e^x - e^{-x})]}{(e^x + e^{-x})^2}$
$\displaystyle \frac{2e^{-x} \cdot 2e^x}{(e^x + e^{-x})^2}$
$\displaystyle \frac{4}{(e^x + e^{-x})^2}$
Ok let's see if I can type this out correctly.
#1) $\displaystyle \frac{1}{x+\sqrt{x^2+1}} \left(1 + \frac{2x}{2 \sqrt{x^2+1}} \right)$
So distribute the first expression through the parenthesis.
$\displaystyle \frac{1}{x+\sqrt{x^2+1}}+\left( \frac{1}{x+\sqrt{x^2+1}} \right) \left(\frac{2x}{2 \sqrt{x^2+1}} \right)$
Multiply the last two expressions.
$\displaystyle \frac{1}{x+\sqrt{x^2+1}}+\frac{2x}{2x \sqrt{x^2+1}+2(x^2+1)}$
The denominator of expression two can be factored with some moving around.
$\displaystyle \frac{1}{x+\sqrt{x^2+1}}+\frac{2x}{(x+\sqrt{x^2+1} )^2-1}$
Now multiply each expression by the conjugate of the denominator to get the radicals gone.
EDIT: I'll keep the post here just because it took me so long, but look above at skeeter's post for a much quicker, better simplification.
$\displaystyle \frac{1}{x + \sqrt{x^2+1}} \left(\frac{x + \sqrt{x^2+1}}{\sqrt{x^2+1}}\right)$
Ok so you can see this easier lets
$\displaystyle (x + \sqrt{x^2+1}) = a$
so it would look like this
$\displaystyle \bigg(\frac{1}{a}\bigg) \bigg(\frac{a}{\sqrt{x^2+1}}\bigg)$