anonymous
  • anonymous
\[\int\frac{ (lnx) ^{2} }{ x }dx\]
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
u= 2ln(x) du = 2/x
anonymous
  • anonymous
then
anonymous
  • anonymous
integral of 1/2*u*du

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

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