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amurciano2
 one year ago
How would you integrate 3x(ln(x))^2
amurciano2
 one year ago
How would you integrate 3x(ln(x))^2

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amurciano2
 one year ago
Best ResponseYou've already chosen the best response.1I know that it is an integration by parts, but I don't know what to set as my u and dv.

amurciano2
 one year ago
Best ResponseYou've already chosen the best response.1I 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?

AravindG
 one year ago
Best ResponseYou've already chosen the best response.1L = logarithmic I = inverse trigonometric A = algebraic T = trigonometric E = exponential

AravindG
 one year ago
Best ResponseYou've already chosen the best response.1I>L>A>T>E That is the order of priority.

amurciano2
 one year ago
Best ResponseYou've already chosen the best response.1would I set (lnx)^2 as my U. does it become part of the log family or exponential ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1give it a try, you would never know if it works or not without trying

AravindG
 one year ago
Best ResponseYou've already chosen the best response.1You have to consider the inside function. f(x)^2 f(x) belongs to?

amurciano2
 one year ago
Best ResponseYou've already chosen the best response.1yes that is why I put lnx as my u

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0lmltfy! \[\int 3x(ln(x))^2 \ dx\]

amurciano2
 one year ago
Best ResponseYou've already chosen the best response.1yes that is why I put lnx as my udw:1440859979280:dw

amurciano2
 one year ago
Best ResponseYou've already chosen the best response.1This is what I did. Does it look correct?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3It is too blur to see but I think for uv, you are ok, for \(\int vdu\) it is not ok to me.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3You let u = (ln(x))^2 > \(du = \dfrac{2lnx}{x}dx\) dv = 3xdx > \(v = \dfrac{3x^2}{2}\) , right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3Hence 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
 one year ago
Best ResponseYou've already chosen the best response.3apply integral by part again for the second term, you can get the answer.

amurciano2
 one year ago
Best ResponseYou've already chosen the best response.1I backtracked and fixed it.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1A 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
 one year ago
Best ResponseYou've already chosen the best response.0Another 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
 one year ago
Best ResponseYou've already chosen the best response.0that was deep @Empty

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0"When you can take the pebble from my hand, it will be time for you to leave."
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