Integrating e^x

Nas

Apr 2010
11
0
I am unsure on how to calculate the following 2 integrals

a) [1/(x^2)]*[e^(2/x)], with respect to xand
b) 1/(e^x + 1), with respect to x

Help with this will be greatly appreciated.
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
I am unsure on how to calculate the following 2 integrals

a) [1/(x^2)]*[e^(2/x)], with respect to xand
b) 1/(e^x + 1), with respect to x

Help with this will be greatly appreciated.
Here you have

a) \(\displaystyle \int{\left(\frac{1}{x^2}\right)e^{\frac{2}{x}}\,dx} = \int{e^{2x^{-1}}\,x^{-2}\,dx}\)

\(\displaystyle = -\frac{1}{2}\int{e^{2x^{-1}}(-2x^{-2})\,dx}\).

Now make the substitution \(\displaystyle u = 2x^{-1}\) so that \(\displaystyle du = -2x^{-2}\,dx\) and the integral becomes

\(\displaystyle -\frac{1}{2}\int{e^u\,du}\)

\(\displaystyle = -\frac{1}{2}e^u + C\)

\(\displaystyle = -\frac{1}{2}e^{2x^{-1}} + C\)

\(\displaystyle =-\frac{1}{2}e^{\frac{2}{x}} + C\).
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
I am unsure on how to calculate the following 2 integrals

a) [1/(x^2)]*[e^(2/x)], with respect to xand
b) 1/(e^x + 1), with respect to x

Help with this will be greatly appreciated.
b) \(\displaystyle \int{\frac{1}{e^x + 1}\,dx} = \int{\frac{e^x}{e^x(e^x + 1)}\,dx}\)

\(\displaystyle = \int{\left[\frac{1}{e^x(e^x + 1)}\right]e^x\,dx}\).

Now make the substitution \(\displaystyle u = e^x + 1\) so that \(\displaystyle e^x = u - 1\) and \(\displaystyle du = e^x\,dx\).

The integral becomes

\(\displaystyle \int{\frac{1}{(u - 1)u}\,du}\).

Now solve this using partial fractions.
 

Nas

Apr 2010
11
0
would the answer for part b be x-log(e^x+1)?
 
Mar 2010
715
381
  1. Let \(\displaystyle u = \dfrac{2}{x}\), then \(\displaystyle \dfrac{du}{dx} = \dfrac{-2}{x^2} \Rightarrow {dx} = -\dfrac{x^2{du}}{2}\). Therefore \(\displaystyle \int\dfrac{1}{x^2}e^{\frac{2}{x}}\;{dx} = -\int\left(\dfrac{e^u}{x^2}\right)\left(\dfrac{x^2}{2}\right)\;{du} = ...\)
  2. Let \(\displaystyle u = e^x+1\), then \(\displaystyle \dfrac{du}{dx} = e^x \Rightarrow {dx} = \dfrac{du}{e^x}.[/Math] Therefore \(\displaystyle \int\dfrac{1}{e^x+1}\;{dx} = \int\dfrac{1}{u(u-1)}\;{du} = ... \)\)\(\displaystyle
    \)
 
Jan 2010
564
242
Kuwait
Another solution for 2:

\(\displaystyle \int \frac{dx}{e^x+1} = \int \frac{1}{e^x+1} \, \frac{e^{-x}}{e^{-x}} \, dx\)

\(\displaystyle =\int \frac{e^{-x}}{e^{-x}+1} \, dx\)

Let \(\displaystyle u=e^{-x}+1\), to get :

\(\displaystyle - \int \frac{du}{u} = -ln|u|+C = -ln(e^{-x}+1) + C\)

\(\displaystyle =ln \left( \frac{1}{e^{-x}+1} \right) + C\)
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, Nas!

The General beat me to the solution . . . *sigh*


\(\displaystyle (b)\;\;\int\frac{dx}{e^x+1} \)

I have an equivalent answer: .\(\displaystyle -\ln\left(1 + e^{-x}\right) + C\)


And this answer can be simplified beyond all recognition . . .


\(\displaystyle -\ln\left(1 + \frac{1}{e^x}\right) + C\;=\;-\ln\left(\frac{e^x+1}{e^x}\right) + C\)

. . . . . . . . . . . . \(\displaystyle =\;-\ln(e^x+1) + \ln(e^x) + C\)

. . . . . . . . . . . . \(\displaystyle =\;-\ln(e^x+1) + x\underbrace{\ln(e)}_{\text{This is 1}} + C\)

. . . . . . . . . . . . \(\displaystyle =\;x - \ln(e^x+1) + C\)

 
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