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#1 2014-08-25 07:05:28

Au101
Member
Registered: 2010-12-01
Posts: 353

Integration by parts question

Having moved on to integration by parts, I've found myself stuck on another question:

I have two problems here. My first is that I get the answer:

Whilst the book gets:

That's no problem, I thought, because:

But when I checked whether this was true in WolframAlpha, it told me that it was false, so I was wondering whether my answer was, in fact, equivalent to the one in the book or not?

My second problem is that I came by this solution by adapting a worked example of

Which I found online.

I did it by setting:

But the problem I have is that I don't really understand why that is the thing to do. In all of the other examples in my book, I've had a product of two functions of x (often f(x) = x). I have then selected one of the functions of x to be u and integrated the other function of x. The result of the integration I then set as v. Here I've got something totally different and the book offers no guidance, unfortunately, so I was wondering if anyone could explain how this works as I may well have similar problems in future.

Last edited by Au101 (2014-08-25 07:05:55)

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#2 2014-08-25 07:56:34

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Integration by parts question

Hi;

To deal with the first part of your question, those two forms are not always equal. You can get Wolfram to say they are equal by entering this:

PowerExpand({-Log[(x - 1)/(x + 1)],Log[(x + 1)/(x -1)]})

You see for some Real values like x >1 those two expressions are the same but for complex values they are not. That is why Alpha says they are not equal. Wolfram is very precise about things. Using PowerExpand forces him to use positive real values only. Then he expands those two logs so you can see they are the same.

Math is about being cunning.

To try integration by parts on that you could start with

now you have two parts like you are used to. I am not saying that is the answer, I am just showing you how you can possibly set it up with a trick of dividing and then multiplying. There are of course other ways to try.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#3 2014-08-25 08:14:40

Au101
Member
Registered: 2010-12-01
Posts: 353

Re: Integration by parts question

I see, I now understand why wolfram gave me the answer it did. I had wondered whether it might be because wolfram uses complex values as well, but until you explained I didn't know how to get it to only use real values of x - thanks! big_smile

As for the integration by parts, I suppose I will just have to do a little more practise so that I can spot what to do. Thanks again smile

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#4 2014-08-25 08:21:38

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Integration by parts question

Look, integration is tough. So tough, it just can not be done most of the time. When you are going through a book the problem is somewhat simplified. If you are in the substitution section you automatically know that you use substitution on the integral. When in the IBP section you have a strong hint that IBP will work, or that repeated use of IBP will work. When you get one of these integrals on your own you will have no clues what to even try. Often, you will try lots of different ideas before you get something that looks promising.

The more you practice the better your intuition will be as to what might work. Good luck and hard work can get you through.

Check this page out:

http://calculus-geometry.hubpages.com/h … ate-Lnx2-1

because wolfram uses complex values as well.

As far as I know it defaults always to C. Many common place simplifications then do not apply.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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