anonymous
  • anonymous
integral of ((lnx)^5)/(5x)dx please help step by step.. it's a practice problem for my exam this coming week.. and I need help reviewing because I've forgotten basically how to do these problems
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
This is a problem you want to solve by using the method of substitution. Got any idea how to do that?
anonymous
  • anonymous
u substitution..? I don't know how to do it.. it confuses me. the only thing i know is u= lnx and du= 1/x ?
anonymous
  • anonymous
First I recommend you to rewrite the integral, like this: \[\large \frac{1}{5}\int (\ln x)^5 \cdot \frac{1}{x}dx \]

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anonymous
  • anonymous
Yes, perfect @monroe17 !! You're pretty close from there.
anonymous
  • anonymous
wait why and how did you rewrite it like that?
anonymous
  • anonymous
Well, first I did factor out the constants, then I just rewrote it to make the substitution more obvious. The rules are just algebraic. (Providing I understood your problem correctly) \[ \large \frac{(\ln x)^5}{5x}= (\ln x)^5 \cdot \frac{1}{5x} \]
anonymous
  • anonymous
question: (this may be a stupid question too) how'd you get 1/5 factored out?
anonymous
  • anonymous
It's not too obvious if you're new to integrals, but it's a general rule of integration, you're integrating in the dimension / direction of \(dx\) therefore all numbers \( \in \mathbb{R} \) can be factored out. This will make your integral look 'easier' Or at least you wont be confused by the numbers anymore. Of course you must multiply them back in at the end. However, to some people this seems like uneasy extra work and they just skip that step, which is perfectly fine too.
anonymous
  • anonymous
did it come from the 1/5x part?
anonymous
  • anonymous
exactly, I just thought the integral would look nicer without the unnecessary 5 in the denominator.
anonymous
  • anonymous
but then why did it stay 1/x ... why wouldn't it be x?
anonymous
  • anonymous
as an example:\[\Large \int \frac{1}{5x}dx = \frac{1}{5} \int \frac{1}{x}dx \]
anonymous
  • anonymous
Just a general law of multiplication \[ \large \frac{a}{b} \cdot \frac{c}{d}= \frac{ac}{bd} \]
anonymous
  • anonymous
oh lol.. okay
anonymous
  • anonymous
Most teachers provide that step, to some students - new to calculus - if you show them the integrals I posted above, they would say that they don't know how to integrate the first one, but the second one. It's more like to put empathy to the general structure of it, most of the time that's all what mathematics is about (-:
anonymous
  • anonymous
oh okayy.. so now do i find the anti derivative?
anonymous
  • anonymous
Use the substitution you have already figured out yourself, if you plug them into correctly you will see that it dramatically simplifies this devilish looking integral.
anonymous
  • anonymous
lol! okay :) give me a second
anonymous
  • anonymous
\[\frac{ 1 }{ 5 }\int\limits_{?}^{?}(u)^5du\] this..?
anonymous
  • anonymous
perfect
anonymous
  • anonymous
You know ho to integrate that right?
anonymous
  • anonymous
uhm, i dont, this is where i get stuck the majority of the time. i dont get this part
anonymous
  • anonymous
would you know hot integrate this expression \[ \large \int x^3dx \] ?
anonymous
  • anonymous
excuse my poor spelling (-; how to *
anonymous
  • anonymous
isn't it just finding the antideriv?
anonymous
  • anonymous
exactly
anonymous
  • anonymous
with the general rule: \[ \Large \int x^ndx = \frac{1}{n+1}x^{n+1}+C\]
anonymous
  • anonymous
You've managed to simplify your integral to exact that form, except that in your case it says \(u\) rather than \(x\)
anonymous
  • anonymous
okay hold on :)
anonymous
  • anonymous
\[\frac{ 1 }{ 5 }\int\limits_{?}^{?}\frac{ 1 }{ 6 }u^6+C\]
anonymous
  • anonymous
exactly, now you can just multiply it out and you're done.
anonymous
  • anonymous
well, after backsubstitution you're done \[ u = \ln x \]
anonymous
  • anonymous
what do you mean multiply it out? the 1/6 ?
anonymous
  • anonymous
by the way, I guess you mean the right thing, but you're spelling is a bit incorrect. You better write it like that: \[ \large \frac{1}{5}\int u^5du = \frac{1}{5}\left( \frac{1}{6}u^6+C\right) \]
anonymous
  • anonymous
see the general rule of integration above again. the RHS doesn't include any integral sign anymore.
anonymous
  • anonymous
oh i get it!
anonymous
  • anonymous
great (-: !
anonymous
  • anonymous
Thank you, you're great at explaining :)
anonymous
  • anonymous
Appreciate it, you're welcome

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