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

1. amurciano2

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

2. AravindG

ILATE

3. 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?

4. AravindG

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

5. AravindG

I>L>A>T>E That is the order of priority.

6. amurciano2

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

7. ganeshie8

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

8. AravindG

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

9. amurciano2

yes that is why I put lnx as my u

10. IrishBoy123

lmltfy! $\int 3x(ln(x))^2 \ dx$

11. amurciano2

yes that is why I put lnx as my u|dw:1440859979280:dw|

12. amurciano2

This is what I did. Does it look correct?

13. Loser66

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

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

15. Loser66

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

16. Loser66

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

17. amurciano2

I backtracked and fixed it.

18. 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.

19. 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 $$dg$$.

20. ikram002p

that was deep @Empty

21. IrishBoy123

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

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