My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree2Thanks
  • 1 Post By v8archie
  • 1 Post By greg1313
Reply
 
LinkBack Thread Tools Display Modes
January 15th, 2015, 04:06 PM   #1
Newbie
 
Joined: Jan 2015
From: West Virginia

Posts: 11
Thanks: 0

Integrals

⌠ (e^x+1)/e^x


Supposed to use u-substitution. When I use u = e^x+1 and du = e^x, I end up with du on the bottom and am not sure how to deal with it. Thanks for the help!
joshbeldon is offline  
 
January 15th, 2015, 04:19 PM   #2
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 6,857
Thanks: 2230

Math Focus: Mainly analysis and algebra
You shouldn't have a $\mathrm d u$ on the bottom. I think that you should have $\mathrm e^x = u-1$ there instead.

However I think you are using the wrong substitution. Try $u=\mathrm e^x$ instead. You will want to multiply both the numerator and the denominator by $\mathrm e^x$ before doing the substitution.

Personally I wouldn't use a substitution at all.
Thanks from topsquark

Last edited by v8archie; January 15th, 2015 at 04:28 PM.
v8archie is offline  
January 15th, 2015, 04:51 PM   #3
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,517
Thanks: 910

Math Focus: Elementary mathematics and beyond
$\displaystyle \int\frac{e^x+1}{e^x}\,dx=\int(e^x+1)e^{-x}\,dx$

Now make the substitution $\displaystyle u=e^{-x}$, integrate and simplify.
Thanks from topsquark
greg1313 is offline  
January 15th, 2015, 05:00 PM   #4
Newbie
 
Joined: Jan 2015
From: West Virginia

Posts: 11
Thanks: 0

When I make u = e^(-x), that leave du = -e^(-x). The remaining number is (e^x - 1), so I'm not sure they cancel out.
joshbeldon is offline  
January 15th, 2015, 05:15 PM   #5
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,517
Thanks: 910

Math Focus: Elementary mathematics and beyond
If $\displaystyle u=e^{-x}$ then $\displaystyle e^x=\frac1u$
greg1313 is offline  
January 16th, 2015, 07:41 AM   #6
Math Team
 
topsquark's Avatar
 
Joined: May 2013
From: The Astral plane

Posts: 1,570
Thanks: 613

Math Focus: Wibbly wobbly timey-wimey stuff.
Or, more directly
$\displaystyle \int \frac{e^x + 1}{e^x}~dx = \int (1 + e^{-x})~dx = \int dx + \int e^{-x}~dx$
by simple division.

-Dan
topsquark is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
integrals



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
integrals bonildo Calculus 10 September 12th, 2014 05:19 PM
integrals Random Variable Calculus 9 January 9th, 2012 04:31 AM
Nine Integrals guynamedluis Complex Analysis 18 January 2nd, 2012 06:46 PM
integrals pluto500 Calculus 5 October 18th, 2011 04:47 PM
def. integrals FreaKariDunk Calculus 4 September 12th, 2011 04:24 AM





Copyright © 2017 My Math Forum. All rights reserved.