Difficult partial fractions and/or difficult integral

Au101 · 2014-08-19 15:05:52

Hi again, so the next question that has me completely stumped is:

I have completed the proof, but am stuck on the next half of the question. Because of what it asked me to prove, I attempted to solve the second integral and my method was to try using the substitution:

Here's what I have:

And here is where I get completely stuck. WolframAlpha tells me that this does have a partial fraction expansion, but I can't for the life of me work out how to arrive at it, since neither:

Nor:

Factorise, meaning that I cannot make either of them zero.

So, either I'm missing a trick and there is something I can do to make these zero, or there's a partial fractions technique I've not learned, or I'm barking up entirely the wrong tree with my substitution.

Any ideas?

bobbym · 2014-08-19 18:00:54

Hi;

Try this here:

now substitute u = 0 , 1, -1, 2 and solve the 4 x 4 simultaneous set of linear equations.

Bob · 2014-08-19 19:06:53

hi Au101,

What's wrong with

Bob

Au101 · 2014-08-20 01:11:33

bob bundy wrote:

hi Au101,
What's wrong with
Bob

I never really knew how to do that, but thanks bob bundy, that way's much easier.

Still, I'm glad I now know how to do the partial fractions - thanks bobbym

Au101 · 2014-08-20 01:41:11

What was your method for solving the four simultaneous equations bobbym? I have now done it, but it took me multiple pages!!

bobbym · 2014-08-20 01:44:07

Hi;

That is best done by a computer but it probably can be done by hand too.

1) Clear out fractions and use the 2nd equation to reduce to a 3 x 3.

Au101 · 2014-08-20 01:55:37

Ta! I can show you how I did it, for the sake of interest, but it's nice to know there wasn't a simple method I completely missed:

To be continued...

Last edited by Au101 (2014-08-20 01:59:29)

Au101 · 2014-08-20 01:58:49

...

bobbym · 2014-08-20 02:02:18

Hi;

We can clean up the original equations to be:

Use b = 1 - d and substitute:

Now use a = -c - 1 to reduce to a 2 x 2

Au101 · 2014-08-20 02:05:03

Ah yes, that would have worked very nicely. Much neater!

bobbym · 2014-08-20 02:06:51

We will get:

You can now use c = d - 1 to reduce to a linear equation.

Au101 · 2014-08-20 02:19:14

Certainly, thanks a lot

bobbym · 2014-08-20 02:20:43

I made a little change to the last post. After you have d, you back substitute into the equations above it and everything will be fine.

Bob · 2014-08-20 04:15:27

Au101 wrote:

I never really knew how to do that, but thanks bob bundy, that way's much easier.

I just thought, why did they ask for that substitution, unless it made the integral easier. So in what way ? Thinks ....... Ah ha!

Bob

Math Is Fun Forum

#1 2014-08-19 15:05:52

Difficult partial fractions and/or difficult integral

#2 2014-08-19 18:00:54

Re: Difficult partial fractions and/or difficult integral

#3 2014-08-19 19:06:53

Re: Difficult partial fractions and/or difficult integral

#4 2014-08-20 01:11:33

Re: Difficult partial fractions and/or difficult integral

#5 2014-08-20 01:41:11

Re: Difficult partial fractions and/or difficult integral

#6 2014-08-20 01:44:07

Re: Difficult partial fractions and/or difficult integral

#7 2014-08-20 01:55:37

Re: Difficult partial fractions and/or difficult integral

#8 2014-08-20 01:58:49

Re: Difficult partial fractions and/or difficult integral

#9 2014-08-20 02:02:18

Re: Difficult partial fractions and/or difficult integral

#10 2014-08-20 02:05:03

Re: Difficult partial fractions and/or difficult integral

#11 2014-08-20 02:06:51

Re: Difficult partial fractions and/or difficult integral

#12 2014-08-20 02:19:14

Re: Difficult partial fractions and/or difficult integral

#13 2014-08-20 02:20:43

Re: Difficult partial fractions and/or difficult integral

#14 2014-08-20 04:15:27

Re: Difficult partial fractions and/or difficult integral

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