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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

I know that it is an integration by parts, but I don't know what to set as my u and dv.
ILATE
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?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

L = logarithmic I = inverse trigonometric A = algebraic T = trigonometric E = exponential
I>L>A>T>E That is the order of priority.
would I set (lnx)^2 as my U. does it become part of the log family or exponential ?
give it a try, you would never know if it works or not without trying
You have to consider the inside function. f(x)^2 f(x) belongs to?
yes that is why I put lnx as my u
lmltfy! \[\int 3x(ln(x))^2 \ dx\]
yes that is why I put lnx as my u|dw:1440859979280:dw|
This is what I did. Does it look correct?
1 Attachment
It is too blur to see but I think for uv, you are ok, for \(\int vdu\) it is not ok to me.
You let u = (ln(x))^2 --> \(du = \dfrac{2lnx}{x}dx\) dv = 3xdx --> \(v = \dfrac{3x^2}{2}\) , right?
Hence the integral becomes: \[\dfrac{3x^2(ln(x))^2}{2} -\int \dfrac{3x^2\cancel{2} lnx}{\cancel{2}x} dx\] \[= thefirstterm -\int 3xlnxdx\]
apply integral by part again for the second term, you can get the answer.
I backtracked and fixed it.
1 Attachment
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.
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 \(dg\).
that was deep @Empty
"When you can take the pebble from my hand, it will be time for you to leave."

Not the answer you are looking for?

Search for more explanations.

Ask your own question