Integrating e^X - e^-x/e^-x+1 dx

  • Context: MHB 
  • Thread starter Thread starter markosheehan
  • Start date Start date
  • Tags Tags
    Dx E^x Integrating
Click For Summary

Discussion Overview

The discussion revolves around the integration of the expression \( \frac{e^x - e^{-x}}{e^{-x} + 1} \) or its alternative forms. Participants explore different methods for simplifying the integral, including factoring and substitution techniques.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in integrating the expression and attempts to eliminate the fraction by multiplying by \( e^x \).
  • Another participant clarifies the expression in question, asking whether it is the first or second form of the integral.
  • A participant confirms they are referring to the first form of the integral.
  • One participant suggests factoring \( e^x \) out of the numerator instead of multiplying, leading to a transformation of the integral into a different form involving \( u = e^{-x} \).
  • Another participant questions the reasoning behind factoring out \( e^x \) when the differential requires \( e^{-x} \).

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to approach the integration, with differing opinions on whether to multiply or factor the expression.

Contextual Notes

The discussion includes unresolved mathematical steps and varying assumptions about the manipulation of the integral.

markosheehan
Messages
133
Reaction score
0
how do you integrate e^X-e^-x/e^-x+1 dx
i am trying multiplying by e^x and trying to make it into no fraction but i am having no luck
 
Physics news on Phys.org
markosheehan said:
how do you integrate e^X-e^-x/e^-x+1 dx
i am trying multiplying by e^x and trying to make it into no fraction but i am having no luck

do you mean integrate

$\dfrac{e^x - e^{-x}}{e^{-x}+1}$ or $e^x - \dfrac{e^{-x}}{e^{-x}+1}$
 
The first one
 
markosheehan said:
how do you integrate e^X-e^-x/e^-x+1 dx
i am trying multiplying by e^x and trying to make it into no fraction but i am having no luck
Instead of multiplying by e^x, I would factor an e^x out of the numerator:
$\int \frac{e^x- e^{-x}}{e^{-x}+ 1}dx= \frac{1- e^{-2x}}{e^{-x}+ 1} e^xdx$

Let $u= e^{-x}$ so $du= -e^{-x}dx$. The integral becomes $-\int\frac{1- u^2}{u+ 1}du$. I presume you know that $1- u^2= (1- u)(1+ u)$.
 
HallsofIvy said:
Instead of multiplying by e^x, I would factor an e^x out of the numerator:
$\int \frac{e^x- e^{-x}}{e^{-x}+ 1}dx= \frac{1- e^{-2x}}{e^{-x}+ 1} e^xdx$

Let $u= e^{-x}$ so $du= -e^{-x}dx$. The integral becomes $-\int\frac{1- u^2}{u+ 1}du$. I presume you know that $1- u^2= (1- u)(1+ u)$.

Why would you factor out e^x to be used in your differential when you need e^(-x)?
 

Similar threads

Undergrad Solving the Difficult Integral ##\int_0^{\infty} x^{n+1} e^{-x} \sin(ax) dx##
  • · Replies 5 ·
Replies
5
Views
3K
High School Straightforward integration…
  • · Replies 27 ·
Replies
27
Views
2K
High School Integration by Parts, an introduction I get confused with
  • · Replies 2 ·
Replies
2
Views
3K
Graduate Multiplying divergent integrals using Hardy fields approach
  • · Replies 3 ·
Replies
3
Views
2K
High School Is this identity containing the Gaussian Integral of any use?
  • · Replies 8 ·
Replies
8
Views
2K
Undergrad Why does the integral of sine of x^2 from - infinity to + infinity diverge?
  • · Replies 4 ·
Replies
4
Views
2K
Undergrad Gaussian integral by differentiating under the integral sign
  • · Replies 19 ·
Replies
19
Views
5K
High School Calculating shaded area: I'm getting discrepancies between methods
  • · Replies 3 ·
Replies
3
Views
3K
MHB 7.1.17 int e^{-\dfrac{x^2}{2}} dx from 0 to infty
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Integration of ##e^{-x^2}## with respect to ##x##
  • · Replies 4 ·
Replies
4
Views
3K