lambchamps
\[\int\frac{ (lnx) ^{2} }{ x }dx\]
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Algebraic!
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u= 2ln(x)
du = 2/x
lambchamps
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then
Algebraic!
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integral of 1/2*u*du
lambchamps
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and
Algebraic!
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u srs?
Algebraic!
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or should I say u^2/2 srs?
lambchamps
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don't get it, because the answer is \[\frac{ 1 }{3 }lnx ^{3}+c\]
lambchamps
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i just don't know how to solve it
Algebraic!
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don't think you can write it like that...
calculusfunctions
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Alright, do you know the substitution rule for integrals? Also do you know the derivatives of logarithm functions? If you do, then we're off to a great start!
lambchamps
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yep
calculusfunctions
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Were you replying to me or Algebraic?
lambchamps
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you
calculusfunctions
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OK, so then when you see\[\int\limits_{}^{}\frac{ \ln x ^{2} }{ x }dx\]What do you see as the first potential step?
Callisto
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Sorry to interrupt, is the question (i) \(\int \frac{lnx^2}{x}dx\) or (ii) \(\int \frac{(lnx)^2}{x}dx\)? They are different...
calculusfunctions
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Do you know your logarithm properties? for example\[\ln x ^{n}=n \ln x\]Do you know this property?
Algebraic!
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yeah it's supposed to be (ln(x))^2
calculusfunctions
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@Callisto, it's the former.
Callisto
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For the first one, I don't think you can get (1/3)(lnx)^3 +C
Algebraic!
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nope:)
lambchamps
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turn it to \[\int\limits_{?}^{?}\frac{ 1 }{ x } \times \ln x ^{2} dx\]
Algebraic!
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that clear it up @lambchamps ?
lambchamps
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the second @Callisto
Callisto
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You see... That's why...
lambchamps
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sorry it is the second guys
Algebraic!
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so it's
u = ln(x)
u^2 = (ln(x))^2
du = 1/x
go to town.
calculusfunctions
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@Callisto, you are right but @lambchamps, Is the question written correctly before we proceed further?
lambchamps
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@calculusfunctions it's the second. based on what @Callisto have mentioned
lambchamps
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@Algebraic! please continue
calculusfunctions
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OK! so then we don't need the logarithm property I proposed earlier.
Algebraic!
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that's it man, plug em in and integrate.
calculusfunctions
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@lambchamps, do you now what the derivative of\[y =\ln x\]is?
Algebraic!
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@calculusfunctions he or she is in calc. 2 so yeah probably. good question though.
lambchamps
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dy = 1/x dx?
calculusfunctions
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Great, so then if\[\int\limits_{}^{}\frac{ (\ln x)^{2} }{ x }dx\]then in order to apply the substitution rule what should u equal?
lambchamps
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\[=\int\limits_{}^{}\frac{ 1 }{ x } \times \ln x ^{2} dx\]
lambchamps
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i mean (ln x)^2
calculusfunctions
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First of all you keep confusing the issue by writing\[\ln x ^{2}\]instead of\[(\ln x)^{2}\]They are not the same!\[(\ln x)^{2}\neq \ln x ^{2}\]Do you understand? So we're not going to get anywhere until this gaffe is resolved.
calculusfunctions
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OH, OK! Sorry, I see you did fix it.
lambchamps
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sorry it's \[(\ln x)^{2}\]
calculusfunctions
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Do you notice that the derivative of ln x is in the integrand? So then what should u equal?
lambchamps
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don't know please do tell
lambchamps
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is it the \[\frac{ (\ln x)^{n+1} }{ n+1 } ?\]
calculusfunctions
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Here are your options. Do you think u should equal a). ln x or b). 1/x c). (ln x)²
calculusfunctions
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Which option do you think? a, b, or c? Just keep in mind that whichever option you choose, it's derivative must be in the integrand.
lambchamps
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the derivative of ln x is 1/x so i choose b
calculusfunctions
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NO! I said that the derivative of the chose option must be in the integrand. NOT the antiderivative of the option must be in the integrand.
calculusfunctions
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So what should u equal?
lambchamps
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a?
calculusfunctions
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Excellent!
lambchamps
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and then
calculusfunctions
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So now\[u =\ln x\]so then\[du =?\]
lambchamps
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1/x
calculusfunctions
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\[du =\frac{ 1 }{ x }dx\]OK?
lambchamps
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ok sorry
calculusfunctions
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So now if you substitute u and du into your integral, what do you have?
calculusfunctions
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@integralsabiti, how is giving the solution helping the student who is trying to learn? I spent all this time trying to teach @lambchamps so that she can then do other similar problems with confidence, and you just came in wasted her time and my effort. NOT COOL!
lambchamps
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@calculusfunctions effort appreciated.. thanks to both of you
integralsabiti
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sorry for interrupting .you may go on
calculusfunctions
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@integralsabiti, thank you, no worries now that I know your intentions were genuine.
calculusfunctions
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@lambchamps, are you still there? I'm still waiting for your response to my last question regarding your problem.
lambchamps
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so I would need to find the dx then substitute the value to it?
mikala1
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ok calculausfunction after your done here would you help me on my problem pleases
calculusfunctions
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So what do you you now have after substituting\[u =\ln x\]and\[du =\frac{ 1 }{ x }dx\]into\[\int\limits_{}^{}\frac{ (\ln x)^{2} }{ x }dx\]
calculusfunctions
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Of course but let's hurry and finish this one first because I have to log out soon.
calculusfunctions
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So what does the integral look like after substitution?
lambchamps
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dx would be xdu?
calculusfunctions
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NO! If \[u =\ln x\]and\[du =\frac{ 1 }{ x }dx\]then\[\int\limits\limits_{}^{}\frac{ (\ln x)^{2} }{ x }dx =\int\limits_{}^{}u ^{2}du\]Do you see how?
lambchamps
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so the 1/x in (ln x)^2/x would be cancelled?
calculusfunctions
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@mikala, I'll help you right after I finish here.
lambchamps
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because of the x.du
lambchamps
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please tell me i'm right
calculusfunctions
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No, the\[\frac{ 1 }{ x }dx\]is being replaced with du because\[du =\frac{ 1 }{ x }dx\]Please tell me you see that.
calculusfunctions
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Cancelled is a poor choice of words.
lambchamps
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ok, but where did the 1/x go? the one below (ln x)^2
lambchamps
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oh i'm sorry didn't see at first
calculusfunctions
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\[\int\limits\limits_{}^{}\frac{ (\ln x)^{2} }{ x }dx\]is exactly the same as\[\int\limits_{}^{}(\frac{ 1 }{ x }∙(\ln x)^{2})dx\]correct?
lambchamps
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yes
lambchamps
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i see the 1/x dx
lambchamps
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after finding that, then is the time to integrate, right?
calculusfunctions
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So I replaced the factor of\[\frac{ 1 }{ x }dx\]with du because\[du =\frac{ 1 }{ x }dx Yes, now you may integrate\[\int\limits_{}^{}u ^{2}du\]
calculusfunctions
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Sorry, I don't know what happened there let me try again.
lambchamps
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ok
calculusfunctions
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Yes, now find the\[\int\limits_{}^{}u ^{2}du\]Can you do that please?
lambchamps
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\[\frac{ 1 }{ 3 }u ^{3} + c\]
calculusfunctions
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Now what's the final step?
lambchamps
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then substitute
lambchamps
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ln x right?
calculusfunctions
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Yes so can you please write the final answer now?
lambchamps
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\[\frac{ (\ln x)^{3} }{ 3 } + C\] ayt?
calculusfunctions
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Perfect!!!
lambchamps
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on more question is the c suppose to capitalized?
lambchamps
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one*
calculusfunctions
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That doesn't matter. C represents a constant. Whether you write c or C or k or K etc. is irrelevant. Just don't use x, y, or z.
calculusfunctions
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The most commonly used ones are c and k.
lambchamps
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thank you very much Sheldon Cooper you're the best
calculusfunctions
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HAHAHA! @lambchamps, thank you very much! "The Big Bang Theory" is my favourite show!
lambchamps
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i bet.. later dude