# Finding the antiderivative (inverse) of a function

f(x)=1+e^x/1-e^x

## Homework 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]

## 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:

## Answers and Replies

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
Try multiplying the top and bottom by $$e^{-x/2}$$.

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

Last edited:
HallsofIvy
Science Advisor
Homework Helper
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)).