Calculus

[AbysmalGuilt]

Ok, so the second one is really [(1/x)(ln|x|)^3]dx, and you want to integrate. Well, first hand I'll give ya the derivative of ln|x|, which is 1/x. And the derivate you're given has 1/x in it, so it has to come from somewhere. Hope this gives you a big enough hint.

For the first one, now that's a pain in the ass. Mmm....there's a section on integration by parts. I think how it goes is say you have an integral you can treat as dudv. What you want to do is udv - the integral of vdu. So what I would do to start is pick your u to be the easiest one to use, while leave in the integral that's giving you a pain which you can integrate alright by itself.

If you want a hint on a proper u, just ask

Edit: Oh, I know how to do this problem now. Alright, in order to integrate ln|x| properly, given vdu, with v being ln|x| and du being x^4, here's a big hint. What's the derivative of (ln|x|)^2. Then take a look at du. Something is counting down as you integrate by parts over and over again

That's my problem, I'm not able to integrate lnx for dv, I know to pick easiest as u.

[Jerov]

Nonono, don't integrate ln|x|, that's not the point behind integration by parts. Treat ln|x| as your v, so the original integral right now is vdu. So now what this will turn into is vu - the integral udv. Try that bud. If it works, then the new integral you create in the difference should be MUCH easier to integrate

[Mil0]

**** math they are lying to you, you will never really use it.

[Chaotix]

what. the. ****.

[Mil0]

Just when the teacher comes to check your homework be like, See i woulda done it but then it hit me I didnt want to do it and id hate to punish my self like that, she'll just call your house and you can handle that with the simple SHUT THE FUUront door

[jcpiaya]

If you follow the simple formula for integration by parts, shown

http://en.wikipedia.org/wiki/Integration_by_parts

You should not use ln(x) as your v because the problem becomes increasingly difficult if you try to integrate ln(x)

By setting your problem with:

u=ln(x)
du=1/x

v=x^5/5
dv=x^4

The problem becomes much easier.

If you want to solve the problem on your own, don't look further.

ln(x)(x^5/5)- integral((x^5/5)(1/x)dx)

ln(x)*(x^5/5)-(x^5/25)

now to check your answer, we use derivative on this and see if you get the original answer.

Since the first portion follows the multiplication rule.

=(1/x)(x^5/5)+ln(x)(x^4)-(x^4/5)
=(x^4/5)-(x^4/5)+ln(x)(x^4)
=ln(x)(x^4)

Hint for the second problem:

Notice that the derivative of ln(x) is 1/x
Notice that the ln(x) is being "powered"

We can justify using substitution rule for this problem.

by setting
u=ln(x), we get
du=1/x

That should be more than enough information so solve the problem.

[Syetactic]

I was going to take that in grade 12...

-goes and changes schedule-

[AbysmalGuilt]

I was going to take that in grade 12...

-goes and changes schedule-

Yeah, this is actually the first year that I don't catch right on to everything we're doing. I'm usually like top of my class, now I'm like on the bottom. Next to my bad attendance. Darn AP Calculus Good thing I dont NEED it, still gonna try to pass that AP Exam, one way or another.
Thanks for all of the suggestions.

[Mr. First Name Basis]

Just when the teacher comes to check your homework be like, See i woulda done it but then it hit me I didnt want to do it and id hate to punish my self like that, she'll just call your house and you can handle that with the simple SHUT THE FUUront door

Yeah. your cool....

There's this new trend going around that's called going somewhere in life. You should try it.

Sorry haven't taken Calculus yet but my sis is so I know a little bit about derivatives (sp?) but yeah that's pretty much it. I'm not really sure how to integrate so yea... I'm taking Calc next year.. taking Pre Calc now.