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Integration of a division

  • Thread starter SYoung
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  • #1
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This has been posted in another, but the wrong, section of the forum. That's why there are already quotes in it.

Hello,

Homework Statement



I don't understand a step in the following integral:

∫(x-1)/(2x+1)dx = ∫(1/2)dx − (3/2)∫1/(2x+1)dx = (1/2)x − (3/4)ln|2x+1| + C

The first step, where you get the 2 integrals ∫(1/2)dx and -(3/2)∫1/(2x+1)dx
Where do (1/2)dx and -(3/2) come from?
And where does (3/4) come from in the last part?

My best guess is that it's done with partial fractions. But even so, I have no clue how.


Thanks in advance,
Young

Homework Equations



From ordinary division … (2x + 1)/2 = x + 1/2, so x - 1 = (2x + 1)/2 - 3/2
Wow that has been a long time! Could you please show it step by step how it's done in this case? I hope I can remember the whole thing if I see it ;)

From the chain rule :smile:
But isn't it supposed to become ∫1/(2x+1)dx? Since the dirative of ln(x) = 1/x?

The Attempt at a Solution



EDIT:
nvm the chainrule, figured it out ;)
∫ -3/(2 (1+2 x)) dx
= -3/2 ∫ 1/(2 x+1) dx
For the integrand 1/(2 x+1), substitute u = 2 x+1 and du = 2 dx:
= -3/4 ∫ 1/u du
The ∫ of 1/u is ln(u):
= -(3 ln(u))/4+C
Substitute back for u = 2 x+1:
= -3/4 ln(2 x+1)+C


I'm still not sure about the 'ordinary division'.


Thanks again,
Young
 

Answers and Replies

  • #2
699
5
It isn't partial fractions. Like one would do 3/4 to obtain 0.75, you can do the same style of division with polynomials. How many times does (2x+1) go into (x-1)?

Set it up like you would do normal long division.
 
  • #3
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It isn't partial fractions. Like one would do 3/4 to obtain 0.75, you can do the same style of division with polynomials. How many times does (2x+1) go into (x-1)?

Set it up like you would do normal long division.
And that's what my problem is.
I forgot how to do long division, it has been 5 years!

That's why I asked if somebody could please show it step by step for this integral.
I hope I can remember how it's done again once I see it.

I think you didn't read the whole first post
 
  • #5
vela
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One trick you can use is this:

x-1 = (1/2)(2)(x-1) = 1/2 (2x-2) = 1/2 (2x+1-3) = 1/2 (2x+1) - 3/2

It's a bit more straightforward than doing polynomial division.
 
  • #6
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So as this tutorial says:
(x-1)/(2x+1):

x/2x = 1/2
then I carry the x and -1 below
change the signs so they cancel out.

so 2x+1 goes 1/2 times into x-1?
That should be the first part of the integral.. looks like I'm dividing the diratives.

But what about the -(3/2)?


By the way, I'm really getting confused right now.
The integration says: ∫(1/2)dx − (3/2)∫1/(2x+1)dx
and you guys say: 1/2 (2x+1) - 3/2

I doesn't make sense since you substract the 3/2, and thus don't have 1/(2x+1) and the 1/2 is also a multiplication, not the sum as in the integration?????????
 
Last edited:
  • #7
699
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With any remainder, you put it over 2x+1.

Thus, (3/2)/(2x+1).

1/2+[(3/2)/(2x+1)]
 
  • #8
vela
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By the way, I'm really getting confused right now.
The integration says: ∫(1/2)dx − (3/2)∫1/(2x+1)dx
and you guys say: 1/2 (2x+1) - 3/2
That's only the numerator. You still have to divide by 2x+1.
 
  • #9
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That's only the numerator. You still have to divide by 2x+1.
ah now I see.. how could I've been so blind <_<
so you give the denominator and the numerator the same values so you can divide them by eachother.
Giving you the 1/2 - (3/2) * 1/(2x+1) in this case!

Thanks a lot! I was about to give up on this one since I'm running out of studytime for my exam tomorrow. heh
 
  • #10
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And that's what my problem is.
I forgot how to do long division, it has been 5 years!

That's why I asked if somebody could please show it step by step for this integral.
I hope I can remember how it's done again once I see it.

I think you didn't read the whole first post
Your comment about this in your first post was so small, it was difficult to read and therefore, easy to skip over. Think twice about using SIZE = "1" tags around stuff...
 
  • #11
vela
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ah now I see.. how could I've been so blind <_<
so you give the denominator and the numerator the same values so you can divide them by each other.
Yup. For simple cases like this, it's easier to just tweak the top to make everything work out. For more complicated cases, though, you'll need to go back to polynomial division, so you should still know how to do that.
 
  • #12
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Also, I would advise reviewing polynomial long division, as in the link to the tutorial that Dustinsfl gave. If you learned it 5 years ago but have forgotten it, it wouldn't hurt to review it.
 
  • #13
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Thanks for the advise and help everybody.
I'll review the polynomial long division in a few days. I've got exams this week so there's no time left for reviewing.

Thanks again and see you around! :smile:
 

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