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I'm studying for a final, and I haven't seen any mention of any problem of this form in class or in my homework. I can't figure out how to go about solving this problem:

$$\int^{e^6}_{1}{\frac{dx}{x(1+\ln(x))}}$$

What I was thinking is:

$$\int{\frac{dx}{x(1+\ln(x))}}+C = \int{\frac{1}{x(1+\ln(x))}dx}+C =ln(ln(x)+1)+C $$

then solve for $$F(e^6) - F(1)$$

But I'm not so sure this is the correct approach. Can someone highlight why the dx is in such an unusual position?

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    $\begingroup$ It is just a space-saving device. $\endgroup$ – André Nicolas May 4 '13 at 3:09
  • $\begingroup$ At least at this level, the $dx$ (and other possible differentials like $dt$, $dy$, etc.) is just a piece of notation that identifies the variable with respect to which you're integrating. In general, you shouldn't worry too much about it. $\endgroup$ – Javier May 4 '13 at 3:12
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The $\,dx\,$ term in the numerator is simply replacing the multiple $1$.

$$\int^{e^6}_{1}{\frac{dx}{x(1+\ln(x))}} = \int^{e^6}_{1}{\frac{1}{x(1+\ln(x))}\,dx}$$

This is "sort of like" when we represent a fraction in one of two ways: $\;\dfrac 35 = \dfrac 15\cdot 3.\;$

Of course, we don't mean to imply that "$\,dx\,$" is a number, per se. But you will often see "$dx$" positioned in the numerator, instead of alongside and to the right of the function to be integrated.

ADDED: The result of your integration: $\;F(x) = \ln(\ln(x) + 1) + C,\;$ is "spot on": now you need to simply evaluate $\;F(e^6) - F(1),\;$ which I trust you can do!

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  • $\begingroup$ That's what I thought. Just making sure! Thank you!!! $\endgroup$ – alvonellos May 4 '13 at 3:06
  • $\begingroup$ I'll mark your solution as an answer as soon as it lets me. $\endgroup$ – alvonellos May 4 '13 at 3:07
  • $\begingroup$ You're welcome! $\endgroup$ – Namaste May 4 '13 at 3:11

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