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amurciano2

  • one year ago

How would you integrate 3x(ln(x))^2

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  1. amurciano2
    • one year ago
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    I know that it is an integration by parts, but I don't know what to set as my u and dv.

  2. AravindG
    • one year ago
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    ILATE

  3. amurciano2
    • one year ago
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    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?

  4. AravindG
    • one year ago
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    L = logarithmic I = inverse trigonometric A = algebraic T = trigonometric E = exponential

  5. AravindG
    • one year ago
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    I>L>A>T>E That is the order of priority.

  6. amurciano2
    • one year ago
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    would I set (lnx)^2 as my U. does it become part of the log family or exponential ?

  7. ganeshie8
    • one year ago
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    give it a try, you would never know if it works or not without trying

  8. AravindG
    • one year ago
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    You have to consider the inside function. f(x)^2 f(x) belongs to?

  9. amurciano2
    • one year ago
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    yes that is why I put lnx as my u

  10. IrishBoy123
    • one year ago
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    lmltfy! \[\int 3x(ln(x))^2 \ dx\]

  11. amurciano2
    • one year ago
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    yes that is why I put lnx as my u|dw:1440859979280:dw|

  12. amurciano2
    • one year ago
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    This is what I did. Does it look correct?

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  13. Loser66
    • one year ago
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    It is too blur to see but I think for uv, you are ok, for \(\int vdu\) it is not ok to me.

  14. Loser66
    • one year ago
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    You let u = (ln(x))^2 --> \(du = \dfrac{2lnx}{x}dx\) dv = 3xdx --> \(v = \dfrac{3x^2}{2}\) , right?

  15. Loser66
    • one year ago
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    Hence the integral becomes: \[\dfrac{3x^2(ln(x))^2}{2} -\int \dfrac{3x^2\cancel{2} lnx}{\cancel{2}x} dx\] \[= thefirstterm -\int 3xlnxdx\]

  16. Loser66
    • one year ago
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    apply integral by part again for the second term, you can get the answer.

  17. amurciano2
    • one year ago
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    I backtracked and fixed it.

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  18. Empty
    • one year ago
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    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.

  19. anonymous
    • one year ago
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    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 \(dg\).

  20. ikram002p
    • one year ago
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    that was deep @Empty

  21. IrishBoy123
    • one year ago
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    "When you can take the pebble from my hand, it will be time for you to leave."

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