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April 27th, 2017, 06:01 PM   #1
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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
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April 27th, 2017, 07:33 PM   #2
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That integral does not converge, so maybe you're working on the wrong problem.

$$14\int\frac{1}{e^{14x/5}-1}\,dx$$

$$u=\frac{14x}{5},\frac{5}{14}\,du=dx$$

$$5\int\frac{1}{e^u-1}\,du$$

$$e^u=t^2,e^u\,du=2t\,dt$$

$$10\int\frac{1}{t(t^2-1)}\,dt=10\int\frac{t}{t^2-1}-\frac1t\,dt=5\log|t^2-1|-10\log|t|+C$$

$$14\int\frac{1}{e^{14x/5}-1}\,dx+7\int\,dx=5\log|e^{14x/5}-1|-7x+C$$

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Last edited by greg1313; April 28th, 2017 at 10:36 AM.
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April 28th, 2017, 12:59 PM   #3
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$\displaystyle \int\frac{14}{e^\frac{14x}{5}-1}+7 dx$

$\displaystyle u=\frac{14x}{5}$
$\displaystyle du=\frac{14}{5}dx$

$\displaystyle \int\frac{14}{e^u-1}*\frac{5}{14} du+7x$

$\displaystyle p=e^u-1$
$\displaystyle dp=e^udu$

$\displaystyle 5\int\frac{1}{p(p+1)}dp+7x$

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.
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Last edited by phrack999; April 28th, 2017 at 01:02 PM.
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April 30th, 2017, 02:20 AM   #4
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@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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