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monroe17
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
This is a problem you want to solve by using the method of substitution. Got any idea how to do that?
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 ?
First I recommend you to rewrite the integral, like this: \[\large \frac{1}{5}\int (\ln x)^5 \cdot \frac{1}{x}dx \]
Yes, perfect @monroe17 !! You're pretty close from there.
wait why and how did you rewrite it like that?
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) \[ \large \frac{(\ln x)^5}{5x}= (\ln x)^5 \cdot \frac{1}{5x} \]
question: (this may be a stupid question too) how'd you get 1/5 factored out?
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.
did it come from the 1/5x part?
exactly, I just thought the integral would look nicer without the unnecessary 5 in the denominator.
but then why did it stay 1/x ... why wouldn't it be x?
as an example:\[\Large \int \frac{1}{5x}dx = \frac{1}{5} \int \frac{1}{x}dx \]
Just a general law of multiplication \[ \large \frac{a}{b} \cdot \frac{c}{d}= \frac{ac}{bd} \]
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 (-:
oh okayy.. so now do i find the anti derivative?
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.
lol! okay :) give me a second
\[\frac{ 1 }{ 5 }\int\limits_{?}^{?}(u)^5du\] this..?
You know ho to integrate that right?
uhm, i dont, this is where i get stuck the majority of the time. i dont get this part
would you know hot integrate this expression \[ \large \int x^3dx \] ?
excuse my poor spelling (-; how to *
isn't it just finding the antideriv?
with the general rule: \[ \Large \int x^ndx = \frac{1}{n+1}x^{n+1}+C\]
You've managed to simplify your integral to exact that form, except that in your case it says \(u\) rather than \(x\)
\[\frac{ 1 }{ 5 }\int\limits_{?}^{?}\frac{ 1 }{ 6 }u^6+C\]
exactly, now you can just multiply it out and you're done.
well, after backsubstitution you're done \[ u = \ln x \]
what do you mean multiply it out? the 1/6 ?
by the way, I guess you mean the right thing, but you're spelling is a bit incorrect. You better write it like that: \[ \large \frac{1}{5}\int u^5du = \frac{1}{5}\left( \frac{1}{6}u^6+C\right) \]
see the general rule of integration above again. the RHS doesn't include any integral sign anymore.
Thank you, you're great at explaining :)
Appreciate it, you're welcome