# Integration

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• Apr 16th 2010, 09:45 PM
TreePerson
Integration
how would you begin to solve a question like this?

• Apr 16th 2010, 09:49 PM
Debsta
Quote:

Originally Posted by TreePerson
how would you begin to solve a question like this?

Start with your second expression and add the rational expressions by getting a common denominator and checking it is the same as the expression you need to integrate.
Then integrate the second expression in 4 parts.
• Apr 16th 2010, 11:18 PM
TreePerson
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 16th 2010, 11:52 PM
sa-ri-ga-ma
Quote:

Originally Posted by TreePerson
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

You can find the int[2x/(x^2+1)^2] by substituting (x^2 + 1) = u.
Then you can find the int[1/(x^2+1)^2] by substituting x = tanθ.
Now try.
• Apr 16th 2010, 11:59 PM
TreePerson
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, 12:11 AM
Debsta
Quote:

Originally Posted by TreePerson
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

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, 01:52 AM
TreePerson
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, 02:08 AM
Debsta
Quote:

Originally Posted by TreePerson
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?

The expression in the second line of your post
• Apr 17th 2010, 02:11 AM
TreePerson
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, 02:14 AM
Debsta
Quote:

Originally Posted by TreePerson
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

Are you finding it by partial fractions?
• Apr 17th 2010, 02:15 AM
Debsta
Quote:

Originally Posted by TreePerson
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

The question doesn't ask you to "prove" how to get the second expression but just to show they are equal. Therefore you could start with either one and turn it into the other.
• Apr 17th 2010, 02:16 AM
TreePerson
Yeah the cover up rule and the equating coefficients one
• Apr 17th 2010, 02:21 AM
Debsta
Quote:

Originally Posted by TreePerson
Yeah the cover up rule and the equating coefficients one

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, 02:23 AM
TreePerson
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, 02:28 AM
Debsta
Quote:

Originally Posted by TreePerson
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

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