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 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!
 January 15th, 2015, 04:19 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,137 Thanks: 2381 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.
 January 15th, 2015, 04:51 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,724 Thanks: 985 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
 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.
 January 15th, 2015, 05:15 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,724 Thanks: 985 Math Focus: Elementary mathematics and beyond If $\displaystyle u=e^{-x}$ then $\displaystyle e^x=\frac1u$
 January 16th, 2015, 07:41 AM #6 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,663 Thanks: 653 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

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