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 April 27th, 2017, 06:01 PM #1 Newbie   Joined: Apr 2017 From: Melbourne Posts: 2 Thanks: 1 Two ways to integrate Integrate with respect to x: _____14_____ + 7 exp^(14x/5)-1 I got (through intg by substitution multiple times): 7x + 5log(modulus( 1 - exp^(-14x/5) )) but because I need the definite integral over (0, b), subbing in 0 gives an undefined integral On integral-calculator and wolfram-alpha though, I saw that you can also integrate to give: 5log(modulus( exp^(14x/5) - 1 )) - 7x which is great because subbing in 0 is not longer undefined somehow How was the second version integrated? Thank you in advance
 April 27th, 2017, 07:33 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond That integral does not converge, so maybe you're working on the wrong problem. $$14\int\frac{1}{e^{14x/5}-1}\,dx$$ $$u=\frac{14x}{5},\frac{5}{14}\,du=dx$$ $$5\int\frac{1}{e^u-1}\,du$$ $$e^u=t^2,e^u\,du=2t\,dt$$ $$10\int\frac{1}{t(t^2-1)}\,dt=10\int\frac{t}{t^2-1}-\frac1t\,dt=5\log|t^2-1|-10\log|t|+C$$ $$14\int\frac{1}{e^{14x/5}-1}\,dx+7\int\,dx=5\log|e^{14x/5}-1|-7x+C$$ Thanks from mippinwa Last edited by greg1313; April 28th, 2017 at 10:36 AM.
 April 28th, 2017, 12:59 PM #3 Member   Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 $\displaystyle \int\frac{14}{e^\frac{14x}{5}-1}+7 dx$ $\displaystyle u=\frac{14x}{5}$ $\displaystyle du=\frac{14}{5}dx$ $\displaystyle \int\frac{14}{e^u-1}*\frac{5}{14} du+7x$ $\displaystyle p=e^u-1$ $\displaystyle dp=e^udu$ $\displaystyle 5\int\frac{1}{p(p+1)}dp+7x$ Partial Fractions $\displaystyle 5\int\frac{1}{p}-\frac{1}{p+1}dp+7x$ $\displaystyle 5(ln(p)-ln(p+1))+7x$ $\displaystyle 5(ln(e^u-1)-u)+7x$ $\displaystyle 5(ln(e^\frac{14x}{5}-1)-\frac{14x}{5})+7x$ $\displaystyle 5ln(e^\frac{14x}{5}-1)-14x+7x$ $\displaystyle 5ln(e^\frac{14x}{5}-1)-7x$ But with an x of zero the log function is going to fail. That being said this should work for any a to b where a>0. Thanks from mippinwa Last edited by phrack999; April 28th, 2017 at 01:02 PM.
 April 30th, 2017, 02:20 AM #4 Newbie   Joined: Apr 2017 From: Melbourne Posts: 2 Thanks: 1 @greg1313 and @phrack999 Thank you both so much for your replies, decided to redo the question and come back and give my thanks when I got the correct result. Turns out I didn't even solve for the first equation correctly which is why I had problems with getting the definite integral... Anyway, I ended up getting a solution that made sense (don't know if its exactly correct yet) but this was certainly quite an exercise for my integrating skills. Once again, thank you very much for helping me Thanks from greg1313

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