Math Is Fun Forum

jubadedo

Member

Registered: 2015-04-05

Posts: 19

Integration by Parts with natural logs

Hi guys, so I'm working on this problem, and I think I'm (hopefully) close to a solution and I'm just messing up when it comes to simplifying everything. Of course, since this is my own homework, I'm not asking for an answer, just simply a hint or telling me what I'm doing wrong . The problem is this:

   

(I don't know if it will show or not but the problem is the integral from 1 to 2 of 5ln(4x)dx)
So I used

then I said

And when I integrated I got the answer

I factored out a 5x for to get

Then I plugged in the bounds and got

The ones cancelled out and then I factored out a 5 to get

I forget how but I thought I could reduce it to

but it was wrong 

Could someone please guide me to the correct answer? Am I integrating wrong or condensing the logarithms incorrectly or something else?

vaquero

Member

Registered: 2015-08-29

Posts: 1

Re: Integration by Parts with natural logs

You put 10ln(8)-1-5ln(4)-1, but it should be 10ln(8)-10-5ln(4)+5. Because ax(ln(ex)-b)-cx(ln(fx)-d)= axln(ex)-abx-cxln(fx)+cdx.
That was some bad distribution of negative and positive signs and also numbers.
That can be simplify to 5(ln(64)-ln(4)-1).
Which gives you 5(ln(16)-1), because ln(a)-ln(b)= ln(a/b)
That is equal to ln(1,048,576)-5, because aln(b)=ln(b^(a))
My instructors will preferred 5(ln(16)-1) as an answer.

Last edited by vaquero (2015-08-29 07:15:42)

Nehushtan

Member

Registered: 2013-03-09

Posts: 957

Re: Integration by Parts with natural logs

[list=*]
[*]

[/*]
[/list]

---

240 books currently added on Goodreads

zetafunc

Moderator

Registered: 2014-05-21

Posts: 2,432

Website

Re: Integration by Parts with natural logs

Everything looks good up to this point:

Then I plugged in the bounds and got

You made a small mistake when plugging in the bounds.