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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

u= 2ln(x)
du = 2/x

- anonymous

then

- anonymous

integral of 1/2*u*du

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

and

- anonymous

u srs?

- anonymous

or should I say u^2/2 srs?

- anonymous

don't get it, because the answer is \[\frac{ 1 }{3 }lnx ^{3}+c\]

- anonymous

i just don't know how to solve it

- anonymous

don't think you can write it like that...

- 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

yep

- calculusfunctions

Were you replying to me or Algebraic?

- anonymous

you

- 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

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

Do you know your logarithm properties? for example\[\ln x ^{n}=n \ln x\]Do you know this property?

- anonymous

yeah it's supposed to be (ln(x))^2

- calculusfunctions

@Callisto, it's the former.

- Callisto

For the first one, I don't think you can get (1/3)(lnx)^3 +C

- anonymous

nope:)

- anonymous

turn it to \[\int\limits_{?}^{?}\frac{ 1 }{ x } \times \ln x ^{2} dx\]

- anonymous

that clear it up @lambchamps ?

- anonymous

the second @Callisto

- Callisto

You see... That's why...

- anonymous

sorry it is the second guys

- anonymous

so it's
u = ln(x)
u^2 = (ln(x))^2
du = 1/x
go to town.

- calculusfunctions

@Callisto, you are right but @lambchamps, Is the question written correctly before we proceed further?

- anonymous

@calculusfunctions it's the second. based on what @Callisto have mentioned

- anonymous

@Algebraic! please continue

- calculusfunctions

OK! so then we don't need the logarithm property I proposed earlier.

- anonymous

that's it man, plug em in and integrate.

- calculusfunctions

@lambchamps, do you now what the derivative of\[y =\ln x\]is?

- anonymous

@calculusfunctions he or she is in calc. 2 so yeah probably. good question though.

- anonymous

dy = 1/x dx?

- 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

\[=\int\limits_{}^{}\frac{ 1 }{ x } \times \ln x ^{2} dx\]

- anonymous

i mean (ln x)^2

- 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

OH, OK! Sorry, I see you did fix it.

- anonymous

sorry it's \[(\ln x)^{2}\]

- calculusfunctions

Do you notice that the derivative of ln x is in the integrand? So then what should u equal?

- anonymous

don't know please do tell

- anonymous

is it the \[\frac{ (\ln x)^{n+1} }{ n+1 } ?\]

- calculusfunctions

Here are your options. Do you think u should equal a). ln x or b). 1/x c). (ln x)Â²

- 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

the derivative of ln x is 1/x so i choose b

- 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

So what should u equal?

- anonymous

a?

- calculusfunctions

Excellent!

- anonymous

and then

- calculusfunctions

So now\[u =\ln x\]so then\[du =?\]

- anonymous

1/x

- calculusfunctions

\[du =\frac{ 1 }{ x }dx\]OK?

- anonymous

ok sorry

- calculusfunctions

So now if you substitute u and du into your integral, what do you have?

- 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

@calculusfunctions effort appreciated.. thanks to both of you

- anonymous

sorry for interrupting .you may go on

- calculusfunctions

@integralsabiti, thank you, no worries now that I know your intentions were genuine.

- calculusfunctions

@lambchamps, are you still there? I'm still waiting for your response to my last question regarding your problem.

- anonymous

so I would need to find the dx then substitute the value to it?

- anonymous

ok calculausfunction after your done here would you help me on my problem pleases

- 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

Of course but let's hurry and finish this one first because I have to log out soon.

- calculusfunctions

So what does the integral look like after substitution?

- anonymous

dx would be xdu?

- 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

so the 1/x in (ln x)^2/x would be cancelled?

- calculusfunctions

@mikala, I'll help you right after I finish here.

- anonymous

because of the x.du

- anonymous

please tell me i'm right

- 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

Cancelled is a poor choice of words.

- anonymous

ok, but where did the 1/x go? the one below (ln x)^2

- anonymous

oh i'm sorry didn't see at first

- calculusfunctions

\[\int\limits\limits_{}^{}\frac{ (\ln x)^{2} }{ x }dx\]is exactly the same as\[\int\limits_{}^{}(\frac{ 1 }{ x }âˆ™(\ln x)^{2})dx\]correct?

- anonymous

yes

- anonymous

i see the 1/x dx

- anonymous

after finding that, then is the time to integrate, right?

- 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

Sorry, I don't know what happened there let me try again.

- anonymous

ok

- calculusfunctions

Yes, now find the\[\int\limits_{}^{}u ^{2}du\]Can you do that please?

- anonymous

\[\frac{ 1 }{ 3 }u ^{3} + c\]

- calculusfunctions

Right!

- calculusfunctions

Now what's the final step?

- anonymous

then substitute

- anonymous

ln x right?

- calculusfunctions

Yes so can you please write the final answer now?

- anonymous

\[\frac{ (\ln x)^{3} }{ 3 } + C\] ayt?

- calculusfunctions

Perfect!!!

- anonymous

on more question is the c suppose to capitalized?

- anonymous

one*

- 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

The most commonly used ones are c and k.

- anonymous

thank you very much Sheldon Cooper you're the best

- calculusfunctions

HAHAHA! @lambchamps, thank you very much! "The Big Bang Theory" is my favourite show!

- anonymous

i bet.. later dude

- calculusfunctions

Later!

Looking for something else?

Not the answer you are looking for? Search for more explanations.