January 15th, 2015, 03:06 PM  #1 
Newbie Joined: Jan 2015 From: West Virginia Posts: 11 Thanks: 0  Integrals
⌠ (e^x+1)/e^x ⌡ Supposed to use usubstitution. 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! 
January 15th, 2015, 03:19 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,937 Thanks: 2265 Math Focus: Mainly analysis and algebra 
You shouldn't have a $\mathrm d u$ on the bottom. I think that you should have $\mathrm e^x = u1$ 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. Last edited by v8archie; January 15th, 2015 at 03:28 PM. 
January 15th, 2015, 03:51 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,572 Thanks: 931 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. 
January 15th, 2015, 04:00 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.

January 15th, 2015, 04:15 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,572 Thanks: 931 Math Focus: Elementary mathematics and beyond 
If $\displaystyle u=e^{x}$ then $\displaystyle e^x=\frac1u$

January 16th, 2015, 06:41 AM  #6 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,570 Thanks: 613 Math Focus: Wibbly wobbly timeywimey 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 

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