amurciano2
  • amurciano2
How would you integrate 3x(ln(x))^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amurciano2
  • amurciano2
I know that it is an integration by parts, but I don't know what to set as my u and dv.
AravindG
  • AravindG
ILATE
amurciano2
  • amurciano2
I know that it is an integration by parts, but I don't know what to set as my u and dv. what does that mean?

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More answers

AravindG
  • AravindG
L = logarithmic I = inverse trigonometric A = algebraic T = trigonometric E = exponential
AravindG
  • AravindG
I>L>A>T>E That is the order of priority.
amurciano2
  • amurciano2
would I set (lnx)^2 as my U. does it become part of the log family or exponential ?
ganeshie8
  • ganeshie8
give it a try, you would never know if it works or not without trying
AravindG
  • AravindG
You have to consider the inside function. f(x)^2 f(x) belongs to?
amurciano2
  • amurciano2
yes that is why I put lnx as my u
IrishBoy123
  • IrishBoy123
lmltfy! \[\int 3x(ln(x))^2 \ dx\]
amurciano2
  • amurciano2
yes that is why I put lnx as my u|dw:1440859979280:dw|
amurciano2
  • amurciano2
This is what I did. Does it look correct?
1 Attachment
Loser66
  • Loser66
It is too blur to see but I think for uv, you are ok, for \(\int vdu\) it is not ok to me.
Loser66
  • Loser66
You let u = (ln(x))^2 --> \(du = \dfrac{2lnx}{x}dx\) dv = 3xdx --> \(v = \dfrac{3x^2}{2}\) , right?
Loser66
  • Loser66
Hence the integral becomes: \[\dfrac{3x^2(ln(x))^2}{2} -\int \dfrac{3x^2\cancel{2} lnx}{\cancel{2}x} dx\] \[= thefirstterm -\int 3xlnxdx\]
Loser66
  • Loser66
apply integral by part again for the second term, you can get the answer.
amurciano2
  • amurciano2
I backtracked and fixed it.
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Empty
  • Empty
A million times what @ganeshie8 said. If you don't know what will happen when you do something, then that means you should probably do it. If it's a dead end then now you know it's a dead end and you have learned something. If it's not a dead end, then you have solved your problem and learned something. This kind of knowledge is also very useful to have on a test because you will know where at least some of your possible paths may lead. It's like they say, the master has failed more than the beginner has ever tried. Think of failure as your step towards becoming the master.
anonymous
  • anonymous
Another thing you can do, although it doesn't prevent you from using IBP, is to substitute \(x=e^u\), so that \(dx=e^u\,du\). This also means \(u=\ln x\), so you have \[3\int x(\ln x)^2\,dx=3\int u^2e^{2u}\,du\] the proceed with IBP by reducing the power of the power term. \[\left.\begin{matrix}f=u^2&&dg=e^{2u}\,du\\ df=2u\,du&&g=\dfrac{1}{2}e^{2u}\end{matrix}\right\}\implies I=\frac{3}{2}u^2e^{2u}-3\int ue^{2u}\,du\] Lather, rinse, repeat. The advantage to making the substitution, in my opinion, is that it transforms the integral into a form that I think makes it more clear which part of the integrand takes on the \(f\) and which takes on the \(dg\).
ikram002p
  • ikram002p
that was deep @Empty
IrishBoy123
  • IrishBoy123
"When you can take the pebble from my hand, it will be time for you to leave."

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