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anonymous
 3 years ago
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
anonymous
 3 years ago
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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is a problem you want to solve by using the method of substitution. Got any idea how to do that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u 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
 3 years ago
Best ResponseYou've already chosen the best response.0First I recommend you to rewrite the integral, like this: \[\large \frac{1}{5}\int (\ln x)^5 \cdot \frac{1}{x}dx \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, perfect @monroe17 !! You're pretty close from there.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait why and how did you rewrite it like that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 3 years ago
Best ResponseYou've already chosen the best response.0question: (this may be a stupid question too) how'd you get 1/5 factored out?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It'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
 3 years ago
Best ResponseYou've already chosen the best response.0did it come from the 1/5x part?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0exactly, I just thought the integral would look nicer without the unnecessary 5 in the denominator.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but then why did it stay 1/x ... why wouldn't it be x?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0as an example:\[\Large \int \frac{1}{5x}dx = \frac{1}{5} \int \frac{1}{x}dx \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Just a general law of multiplication \[ \large \frac{a}{b} \cdot \frac{c}{d}= \frac{ac}{bd} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Most 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
 3 years ago
Best ResponseYou've already chosen the best response.0oh okayy.. so now do i find the anti derivative?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Use 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
 3 years ago
Best ResponseYou've already chosen the best response.0lol! okay :) give me a second

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ 5 }\int\limits_{?}^{?}(u)^5du\] this..?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You know ho to integrate that right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0uhm, i dont, this is where i get stuck the majority of the time. i dont get this part

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0would you know hot integrate this expression \[ \large \int x^3dx \] ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0excuse my poor spelling (; how to *

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0isn't it just finding the antideriv?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0with the general rule: \[ \Large \int x^ndx = \frac{1}{n+1}x^{n+1}+C\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You've managed to simplify your integral to exact that form, except that in your case it says \(u\) rather than \(x\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ 5 }\int\limits_{?}^{?}\frac{ 1 }{ 6 }u^6+C\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0exactly, now you can just multiply it out and you're done.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well, after backsubstitution you're done \[ u = \ln x \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what do you mean multiply it out? the 1/6 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by 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
 3 years ago
Best ResponseYou've already chosen the best response.0see the general rule of integration above again. the RHS doesn't include any integral sign anymore.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you, you're great at explaining :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Appreciate it, you're welcome
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