how would you begin to solve a question like this?

http://www.mypicx.com/uploadimg/1384...04172010_1.png

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- Apr 16th 2010, 10:45 PMTreePersonIntegration
how would you begin to solve a question like this?

http://www.mypicx.com/uploadimg/1384...04172010_1.png - Apr 16th 2010, 10:49 PMDebsta
- Apr 17th 2010, 12:18 AMTreePerson
ok i'm kind of getting there but i dunno how to get the numerator of this part?... could you or someone please show me how? http://www.mypicx.com/uploadimg/1243...04172010_1.png

- Apr 17th 2010, 12:52 AMsa-ri-ga-ma
- Apr 17th 2010, 12:59 AMTreePerson
thanks but na i don't mean how to intergrate it but how to get it in the first place. i don't know how to come up with that numerator

- Apr 17th 2010, 01:11 AMDebsta
Are you starting wiht your second expression (the one made up of 4 rational expressions added together) and getting them over a common denominator as I suggested earlier? Or starting with your first expression and trying to turn it into your second? The first way I suggested is so much easier.

- Apr 17th 2010, 02:52 AMTreePerson
sorry i don't quite know what your getting at, could you explain in more detail or better yet answer it with working! which is the second expression?

- Apr 17th 2010, 03:08 AMDebsta
- Apr 17th 2010, 03:11 AMTreePerson
oh lol i see what your talking about, but the question asked to prove how to get the second expression, so yes I'm starting with my first expression to find the second one. would you know how to do that? I can't get the last of the four parts

- Apr 17th 2010, 03:14 AMDebsta
- Apr 17th 2010, 03:15 AMDebsta
- Apr 17th 2010, 03:16 AMTreePerson
Yeah the cover up rule and the equating coefficients one

- Apr 17th 2010, 03:21 AMDebsta
I haven't heard it called that before....but anyway you'll need a term like (Ax+B)/(x^2+1) and also (Cx+D)/(x^2+1)^2.

Does that make sense to you?

I still think it is sooooo much easier to start with the second one and add terms by getting a common denominator. It would be different if you weren't given the second form and had to do it your way. - Apr 17th 2010, 03:23 AMTreePerson
oh yeah lol you are right, but I'm still curious on how to get from 1 to 2. The exercises I have been doing were based on doing it that way. that's why I didn't think of doing it the other way

- Apr 17th 2010, 03:28 AMDebsta
OK so you'd set it up as:

A/x + B/(x-1) + C/(x-1)^2 + D/(x-1)^3 + (Ex+F)/(x^2+1) + (Gx+H)/(x^2+1)^2

Add it all up and equate your coefficients.

Remember to account for repeated factors on the bottom.

Also remember to put a polynomial of one less degree on the top.