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 integralcalculator and wolframalpha 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. $$14\int\frac{1}{e^{14x/5}1}\,dx$$ $$u=\frac{14x}{5},\frac{5}{14}\,du=dx$$ $$5\int\frac{1}{e^u1}\,du$$ $$e^u=t^2,e^u\,du=2t\,dt$$ $$10\int\frac{1}{t(t^21)}\,dt=10\int\frac{t}{t^21}\frac1t\,dt=5\logt^2110\logt+C$$ $$14\int\frac{1}{e^{14x/5}1}\,dx+7\int\,dx=5\loge^{14x/5}17x+C$$ 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 
$\displaystyle \int\frac{14}{e^\frac{14x}{5}1}+7 dx$ $\displaystyle u=\frac{14x}{5}$ $\displaystyle du=\frac{14}{5}dx$ $\displaystyle \int\frac{14}{e^u1}*\frac{5}{14} du+7x$ $\displaystyle p=e^u1$ $\displaystyle dp=e^udu$ $\displaystyle 5\int\frac{1}{p(p+1)}dp+7x$ Partial Fractions $\displaystyle 5\int\frac{1}{p}\frac{1}{p+1}dp+7x$ $\displaystyle 5(ln(p)ln(p+1))+7x$ $\displaystyle 5(ln(e^u1)u)+7x$ $\displaystyle 5(ln(e^\frac{14x}{5}1)\frac{14x}{5})+7x$ $\displaystyle 5ln(e^\frac{14x}{5}1)14x+7x$ $\displaystyle 5ln(e^\frac{14x}{5}1)7x$ 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. 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 

Tags 
exponential, integrate, logarithm, ways 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
How many ways ?  jbergin  Probability and Statistics  1  December 15th, 2014 02:32 PM 
How many ways...  need2knw  Number Theory  5  August 18th, 2013 09:06 PM 
Ways to 67  daninc55  Number Theory  9  March 18th, 2011 04:35 AM 
Integrate in as many different ways as possible.  The Chaz  Calculus  10  March 4th, 2011 01:18 PM 