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\[\int\frac{ (lnx) ^{2} }{ x }dx\]

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u= 2ln(x) du = 2/x
integral of 1/2*u*du

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Other answers:

u srs?
or should I say u^2/2 srs?
don't get it, because the answer is \[\frac{ 1 }{3 }lnx ^{3}+c\]
i just don't know how to solve it
don't think you can write it like that...
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!
Were you replying to me or Algebraic?
OK, so then when you see\[\int\limits_{}^{}\frac{ \ln x ^{2} }{ x }dx\]What do you see as the first potential step?
Sorry to interrupt, is the question (i) \(\int \frac{lnx^2}{x}dx\) or (ii) \(\int \frac{(lnx)^2}{x}dx\)? They are different...
Do you know your logarithm properties? for example\[\ln x ^{n}=n \ln x\]Do you know this property?
yeah it's supposed to be (ln(x))^2
@Callisto, it's the former.
For the first one, I don't think you can get (1/3)(lnx)^3 +C
turn it to \[\int\limits_{?}^{?}\frac{ 1 }{ x } \times \ln x ^{2} dx\]
that clear it up @lambchamps ?
the second @Callisto
You see... That's why...
sorry it is the second guys
so it's u = ln(x) u^2 = (ln(x))^2 du = 1/x go to town.
@Callisto, you are right but @lambchamps, Is the question written correctly before we proceed further?
@calculusfunctions it's the second. based on what @Callisto have mentioned
@Algebraic! please continue
OK! so then we don't need the logarithm property I proposed earlier.
that's it man, plug em in and integrate.
@lambchamps, do you now what the derivative of\[y =\ln x\]is?
@calculusfunctions he or she is in calc. 2 so yeah probably. good question though.
dy = 1/x dx?
Great, so then if\[\int\limits_{}^{}\frac{ (\ln x)^{2} }{ x }dx\]then in order to apply the substitution rule what should u equal?
\[=\int\limits_{}^{}\frac{ 1 }{ x } \times \ln x ^{2} dx\]
i mean (ln x)^2
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.
OH, OK! Sorry, I see you did fix it.
sorry it's \[(\ln x)^{2}\]
Do you notice that the derivative of ln x is in the integrand? So then what should u equal?
don't know please do tell
is it the \[\frac{ (\ln x)^{n+1} }{ n+1 } ?\]
Here are your options. Do you think u should equal a). ln x or b). 1/x c). (ln x)²
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.
the derivative of ln x is 1/x so i choose b
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.
So what should u equal?
and then
So now\[u =\ln x\]so then\[du =?\]
\[du =\frac{ 1 }{ x }dx\]OK?
ok sorry
So now if you substitute u and du into your integral, what do you have?
@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!
@calculusfunctions effort appreciated.. thanks to both of you
sorry for interrupting .you may go on
@integralsabiti, thank you, no worries now that I know your intentions were genuine.
@lambchamps, are you still there? I'm still waiting for your response to my last question regarding your problem.
so I would need to find the dx then substitute the value to it?
ok calculausfunction after your done here would you help me on my problem pleases
So what do you you now have after substituting\[u =\ln x\]and\[du =\frac{ 1 }{ x }dx\]into\[\int\limits_{}^{}\frac{ (\ln x)^{2} }{ x }dx\]
Of course but let's hurry and finish this one first because I have to log out soon.
So what does the integral look like after substitution?
dx would be xdu?
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?
so the 1/x in (ln x)^2/x would be cancelled?
@mikala, I'll help you right after I finish here.
because of the x.du
please tell me i'm right
No, the\[\frac{ 1 }{ x }dx\]is being replaced with du because\[du =\frac{ 1 }{ x }dx\]Please tell me you see that.
Cancelled is a poor choice of words.
ok, but where did the 1/x go? the one below (ln x)^2
oh i'm sorry didn't see at first
\[\int\limits\limits_{}^{}\frac{ (\ln x)^{2} }{ x }dx\]is exactly the same as\[\int\limits_{}^{}(\frac{ 1 }{ x }∙(\ln x)^{2})dx\]correct?
i see the 1/x dx
after finding that, then is the time to integrate, right?
So I replaced the factor of\[\frac{ 1 }{ x }dx\]with du because\[du =\frac{ 1 }{ x }dx Yes, now you may integrate\[\int\limits_{}^{}u ^{2}du\]
Sorry, I don't know what happened there let me try again.
Yes, now find the\[\int\limits_{}^{}u ^{2}du\]Can you do that please?
\[\frac{ 1 }{ 3 }u ^{3} + c\]
Now what's the final step?
then substitute
ln x right?
Yes so can you please write the final answer now?
\[\frac{ (\ln x)^{3} }{ 3 } + C\] ayt?
on more question is the c suppose to capitalized?
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.
The most commonly used ones are c and k.
thank you very much Sheldon Cooper you're the best
HAHAHA! @lambchamps, thank you very much! "The Big Bang Theory" is my favourite show!
i bet.. later dude

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