Finding the antiderivative (inverse) of a function

1. Jan 4, 2010

Ohmar

1. The problem statement, all variables and given/known data
f(x)=1+e^x/1-e^x

2. Relevant equations
truth is, I don't even know how to approach this, i know i have to swap the variables but now I'm all confused because a friend told me to do this, in this particular case
f(x)=[1+(e^f)-1^x]/[1-(e^f)-1^x]

3. The attempt at a solution
i don't knolw where this -1^x came from, i would give it a shot at how to resolve for x but I really don't know how to get it down from that position as an exponent that is, any hint or advice is truly welcomed.

edit: forget the antiderivative term used in the subject i dont know why i confused those two terms, i just dont know how to delete it now =P

Last edited: Jan 4, 2010
2. Jan 4, 2010

vela

Staff Emeritus
Try multiplying the top and bottom by $$e^{-x/2}$$.

Or try u=$$e^x$$, then partial fractions.

Last edited: Jan 4, 2010
3. Jan 4, 2010

HallsofIvy

Staff Emeritus
Frankly, it sounds to me like you are completely out of your depth. Are you sure you belong in this course? I am particularly concerned that you say titled this "finding the anti-derivative (inverse) of a function". Typically people learn about "inverse functions" in a course at least a semester or a year before the hear about "antiderivatives".
"Finding the anti-derivative" and "finding the inverse" of a function are NOT the same thing at all.

To find the inverse function to f(x)= (1+e^x)/(1-e^x) (Please, please, please, use parentheses! What you wrote was, correctly 1+ (e^x/1)- e^x but I am sure you did not mean that.) I would first write y= (1+e^x)/(1-e^x) (because "y" is simpler to write than "f(x)") and, as you said, "swap" x and y: x= (1+ e^y)/(1- e^y). Now solve for y. Multiply on both sides by 1- e^y to get x(1- e^y)= x- xe^y= 1+ e^y. Now add xe^y to both sides and subtract 1 from both sides to get (1+ x)e^y= x- 1. Divide both sides by x+ 1: e^y= (x-1)/(x+1). Finally, we can solve for y by doing the "inverse function" to the exponential, the natural logarithm: y= ln((x-1)/(x+1)).