Question

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

Answer 1

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

Answer 2

ILATE

Answer 3

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?

Answer 4

L = logarithmic
I = inverse trigonometric
A = algebraic
T = trigonometric
E = exponential

Answer 5

I>L>A>T>E

That is the order of priority.

Answer 6

would I set (lnx)^2 as my U. does it become part of the log family or exponential ?

Answer 7

give it a try, you would never know if it works or not without trying

Answer 8

You have to consider the inside function. f(x)^2
f(x) belongs to?

Answer 9

yes that is why I put lnx as my u

Answer 10

lmltfy!

\[\int 3x(ln(x))^2 \ dx\]

Answer 11

yes that is why I put lnx as my u

Answer 12

This is what I did. Does it look correct?

Answer 13

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

Answer 14

You let u = (ln(x))^2 --> \(du = \dfrac{2lnx}{x}dx\)
dv = 3xdx --> \(v = \dfrac{3x^2}{2}\) , right?

Answer 15

Hence the integral becomes:
\[\dfrac{3x^2(ln(x))^2}{2} -\int \dfrac{3x^2\cancel{2} lnx}{\cancel{2}x} dx\]
\[= thefirstterm -\int 3xlnxdx\]

Answer 16

apply integral by part again for the second term, you can get the answer.

Answer 17

I backtracked and fixed it.

Answer 18

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.

Answer 19

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 \(dg\).

Answer 20

that was deep @Empty

Answer 21

"When you can take the pebble from my hand, it will be time for you to leave."