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I would like to evaluate the following definite integral: $$\int_{2}^{4} \frac {dx} {x \ln x}\ .$$


My working

Let $u = \frac {1} {\ln x}$ and $v' = \frac {1} {x}$

$\implies u' = -\frac {1} {x (\ln x)^2}$ and $v = \ln x$

$\therefore \int_{2}^{4} \frac {dx} {x \ln x} = 1 + \int_{2}^{4} \frac {dx} {x \ln x}$

$\implies 0 = 1$ (say what?)


Answer

$\int_{2}^{4} \frac {dx} {x \ln x} = \ln 2$


When I use the substitution $u = \ln x$ and proceed, I do arrive at the answer, but that is trivial so I am not here to discuss that. What I am here to discuss, however, is my working when I use integration by parts. I seem to have gone wrong somewhere, which I find very intriguing. I assume I must have been careless, but I have been doing calculus all day, so perhaps my mind is fatigued. Even worse, is there some inherent misunderstanding in my concept of integration by parts? I will be very grateful if anyone can point out where I have gone wrong :)


Edit

Following the answers given, it seems I did have a conceptual misunderstanding about integration by parts and it turns out that integration by parts cannot be used to solve this particular integral! Today, I have also found out that integration by parts cannot solve all integrals!

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Integration by parts states $$\int_2^4 uv' dx = uv\big|_2^4 - \int_2^4 u'v dx$$

We have $uv=1$. Hence $uv\big|_2^4 = 0$.

The equation becomes $$\int_{2}^{4} \frac {dx} {x \ln x} = 0 + \int_{2}^{4} \frac {dx} {x \ln x}$$

and no conclusion can be drawn.

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    $\begingroup$ I see. I thought that my $uv$ was independent of $x$, so there was no need to apply the limits to it. $\endgroup$ – Ethan Mark Oct 22 at 15:28
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    $\begingroup$ @EthanMark $uv$ is still a function of $x$. It's just that $uv=x^0$. So $\endgroup$ – Joe Oct 22 at 15:29
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    $\begingroup$ On the contrary, since $uv$ was independent of $x$, applying the limits will always give zero. $\endgroup$ – player3236 Oct 22 at 15:30
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    $\begingroup$ In case my comment wasn't explicit enough, we are doing this: $$x^0\big|_2^4=4^0-2^0=1-1=0$$. $\endgroup$ – Joe Oct 22 at 16:04
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We have $$\int_{2}^{4}\frac{dx}{x\ln(x)}=\big[1\big]_{x=2}^{x=4}+\int_{2}^{4}\frac{dx}{x\ln(x)}=1-1+\int_{2}^{4}\frac{dx}{x\ln(x)}=\int_{2}^{4}\frac{dx}{x\ln(x)}$$ so we don't get anywhere using integration by parts.

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    $\begingroup$ Ah. I thought that my $uv$ was independent of $x$, so there was no need to apply the limits to it. Thank you for your answer still, but since the other answer came just before yours, I will be accepting his :) I still upvoted your answer though! $\endgroup$ – Ethan Mark Oct 22 at 15:29
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    $\begingroup$ Yeah no worries. Also $\int_{2}^{4} \frac {dx} {x \ln x} = 1 + \int_{2}^{4} \frac {dx} {x \ln x}$ gives $0=1$ which isn't true. $\endgroup$ – Äres Oct 22 at 15:32
  • $\begingroup$ I was always taught that, when in doubt, resort to integration by parts as "it solves all integrals". So that is not true then? Integration by parts does fail in some instances, such as this? $\endgroup$ – Ethan Mark Oct 22 at 15:39
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    $\begingroup$ Usually you would try to do a substitution first, and if it fails then try to integrate by parts (see here). $\endgroup$ – Äres Oct 22 at 15:43
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    $\begingroup$ @EthanMark Correctly applying integration by parts, as Äres did, does not result in false statements such as $0=1$. It can however lead to useless results. A bit like when you are trying to solve the equation $x+2y=5$ by using the fact that $x=5-2y$: this means that $(5-2y)+2y=5$ so $5=5$. Now, $5=5$ is true, but did that help us? $\endgroup$ – Joe Oct 22 at 15:44
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$$I=\int_2^4\frac{dx}{x\ln x}$$ $$u=\ln(x)\Rightarrow dx=xdu,\,u\in[\ln(2),\ln(4)]$$ $$I=\int_{\ln(2)}^{\ln(4)}\frac{xdu}{xu}=\int_{\ln(2)}^{\ln(4)}\frac{1}{u}du=\left[\ln(u)\right]_{\ln2}^{\ln4}=\ln(\ln4)-\ln(\ln2)=\ln(2\ln2)-\ln(\ln2)$$ $$=\ln(2)+\ln(\ln2)-\ln(\ln2)=\ln(2)$$ This is really the easiest way to do it, no conclusion can be drawn purely from IBP

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