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

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- anonymous

This is a problem you want to solve by using the method of substitution. Got any idea how to do that?

- 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

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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## More answers

- anonymous

Yes, perfect @monroe17 !! You're pretty close from there.

- anonymous

wait why and how did you rewrite it like that?

- 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

question: (this may be a stupid question too) how'd you get 1/5 factored out?

- 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

did it come from the 1/5x part?

- anonymous

exactly, I just thought the integral would look nicer without the unnecessary 5 in the denominator.

- anonymous

but then why did it stay 1/x ... why wouldn't it be x?

- anonymous

as an example:\[\Large \int \frac{1}{5x}dx = \frac{1}{5} \int \frac{1}{x}dx \]

- anonymous

Just a general law of multiplication
\[ \large \frac{a}{b} \cdot \frac{c}{d}= \frac{ac}{bd} \]

- anonymous

oh lol.. okay

- 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

oh okayy.. so now do i find the anti derivative?

- 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

lol! okay :) give me a second

- anonymous

\[\frac{ 1 }{ 5 }\int\limits_{?}^{?}(u)^5du\] this..?

- anonymous

perfect

- anonymous

You know ho to integrate that right?

- anonymous

uhm, i dont, this is where i get stuck the majority of the time. i dont get this part

- anonymous

would you know hot integrate this expression
\[ \large \int x^3dx \] ?

- anonymous

excuse my poor spelling (-; how to *

- anonymous

isn't it just finding the antideriv?

- anonymous

exactly

- anonymous

with the general rule:
\[ \Large \int x^ndx = \frac{1}{n+1}x^{n+1}+C\]

- anonymous

You've managed to simplify your integral to exact that form, except that in your case it says \(u\) rather than \(x\)

- anonymous

okay hold on :)

- anonymous

\[\frac{ 1 }{ 5 }\int\limits_{?}^{?}\frac{ 1 }{ 6 }u^6+C\]

- anonymous

exactly, now you can just multiply it out and you're done.

- anonymous

well, after backsubstitution you're done
\[ u = \ln x \]

- anonymous

what do you mean multiply it out? the 1/6 ?

- 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

see the general rule of integration above again. the RHS doesn't include any integral sign anymore.

- anonymous

oh i get it!

- anonymous

great (-: !

- anonymous

Thank you, you're great at explaining :)

- anonymous

Appreciate it, you're welcome

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