Given two functions $f(x) = \frac{1}{2}(e^x+e^{-x})$ and $g(x) = \frac{1}{2}(e^x - e^{-x})$, calculate

$$ \int\frac{g(x)}{f(x)} \, dx. $$

Observing that $f'(x) = g(x)$, this is very easy to do. Just take $u = f(x)$ and therefore $du = f'(x)\,dx$. Now we have

$$ \int \frac{f'(x)}{f(x)} \, dx = \int \frac{du}{u} = \ln |u| + C = \ln\left|\frac{1}{2}(e^x+e^{-x})\right|+C. $$

However, this solution is wrong. The correct one is $\ln|e^x+e^{-x}| + C$. I see how we could get that, just plug in $g(x)$ and $f(x)$ directly, and we get

$$ \int \frac{\frac{1}{2}(e^x - e^{-x})}{\frac{1}{2}(e^x+e^{-x})} \, dx. $$

Cancelling the $\frac{1}{2}$'s and taking $u = e^x+e^{-x}$ and $du = (e^x-e^{-x}) dx$ gives us

$$ \int \frac{du}{u} = \ln|u| + C = \ln|e^x+e^{-x}| + C. $$

I can't see where I made a mistake in my approach, and I'd be really grateful if you pointed it out.

  • 1
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    – Shaun
    May 22, 2020 at 18:47
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    $\begingroup$ I feel terrible mentioning it because I know you're only trying to be polite but could you remove the religious elements at the end? There are likely many faiths none should be given a platform $\endgroup$
    – Karl
    May 22, 2020 at 19:10
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    $\begingroup$ @Karl With all due respect, I will not remove it. It is very good to try to propagate the Faith in all contexts of life. The mods may remove it, but hey, I did my best. Not to mention that my right to free speech will be trampled on in this instance. $\endgroup$
    – God bless
    May 22, 2020 at 19:29
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    $\begingroup$ I ment no offence honestly. I'd just rather stick to maths. $\endgroup$
    – Karl
    May 22, 2020 at 19:38
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    $\begingroup$ May the Flying Spaghetti Monster support you in all of your pasta-related endeavours. $\endgroup$
    – TonyK
    Jul 16, 2020 at 13:18

2 Answers 2


Note that if $K$ is a constant, $\ln|K\cdot h(x)|= \ln|K|+\ln|h(x)|$. In other words, the $1/2$ in the log can get 'absorbed' into the $+C$ term.

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    $\begingroup$ Aaaaah, I see! But then that means that my solution $\ln|\frac{1}{2}(e^x + e^{-x})| + C$ doesn't by itself (with $C = 0$) calculate the area under the curve $h(x) = \frac{f(x)}{g(x)}$? $\endgroup$
    – God bless
    May 22, 2020 at 19:00
  • $\begingroup$ It seems $f$ and $g$ are reversed in your comment. It should, in the sense that $$ \int _{0}^x \frac{g(t)}{f(t)}\,dt = \left.\ln|\frac{1}{2}(e^t + e^{-t})|+C\right|_{0}^x $$ $$ = \left(\ln|\frac{1}{2}(e^x + e^{-x})|+C\right)-\left(\ln|\frac{1}{2}(1+1)|+C\right)=\ln|\frac{1}{2}(e^x + e^{-x})| $$The $1/2$ is essential to make sure, starting at $x=0$, the area is $0$; otherwise, you'll be off by a constant. $\endgroup$
    – Integrand
    May 22, 2020 at 19:25
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    $\begingroup$ Yes, $f$ and $g$ are reversed. Sorry about that and thanks for the insight. $\endgroup$
    – God bless
    May 22, 2020 at 19:32

Using the identity: $ \log (\frac{a}{b})=\log(a)- \log(b)$

$\ln \frac{1}{2}(e^x+e^{-x})+C$
$=\ln (e^x+e^{-x})-\ln 2+C $
$=\ln (e^x+e^{-x})+C{'} $ (as $\ln 2$ is constant)


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