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Integrate x/(2x+1) dx

  • Thread starter srfriggen
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  • #1
288
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Homework Statement



[tex]\int[/tex] x/(2x+1) dx

Homework Equations





The Attempt at a Solution



I tried factoring out the 2 on the bottom to get 1/2 [tex]\int[/tex] x/(x+1/2). then I put 1/2 [tex]\int[/tex] x+(1/2)-(1/2)/(x+1/2) dx.

from that i had, 1/2 [tex]\int[/tex]dx - 1/2[tex]\int[/tex]dx/(x+1/2)

finally x/2 - 1/4 ln lx+1/2l + C


My calculator keeps telling me it should be x/2 - 1/4 ln l2x+1l + C and every time try using a different trick i get a different answer. i'm sure i need to add 0, a-a, to get the answer but I can't figure out how to make it work.

If someone could please help me out it and show the steps as to where i went wrong it would be much appreciated. Been racking my brain over this seemingly simple problem. just can't seem to make it work.

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Hurkyl
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finally x/2 - 1/4 ln lx+1/2l + C


My calculator keeps telling me it should be x/2 - 1/4 ln l2x+1l + C and every time try using a different trick i get a different answer.
Wait, why do you think those are different?
 
  • #3
Hurkyl
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P.S. there is an easy way to check your answer, isn't there?
 
  • #4
CompuChip
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Wait, why do you think those are different?
They are, unless the C in one expression is not necessarily the C in the other expression :P
 
  • #5
288
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Hey all, thanks for responding so quickly. I was able to figure out the problem. It was a simple issue of order of operations. When I factored out the 1/2 I didn't realize I have to basically wait till the end of the problem (when all within the brackets [well i wasn't using brackets, that was the problem] is simplified. That made everything work perfectly. whether factoring out a 1/2, or substituting x=u/2, it was all just a simple mistake. Thank god i noticed it now before i went on to my test! i only wonder how many times this mistake has been screwing me in the past!

Also, I knew the answers were different because I graphed them and they were off slightly.

ahh, off to bed.
 
  • #6
Hurkyl
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Also, I knew the answers were different because I graphed them and they were off slightly.
You didn't graph the answers. You graphed one element from each family of answers. Those two particular functions were different, but why do you think the two families were different?
 
  • #7
288
3
You didn't graph the answers. You graphed one element from each family of answers. Those two particular functions were different, but why do you think the two families were different?
Probably because this is all new to me:) I'm not sure I follow. Are you talking about how each solvable indefinite integral has an infinite number of answers all dependent on "+C"? (The graphs were identical except for a small shift in the y-direction).

If that is so then does it matter when i distribute the 1/2 in the front? If I wait until the end of the problem I get the calculator answer. If I distribute it earlier I get my answer.

are the two answers equal except for the constant +C? If so, how would I manipulate the constant to have them be identical?

If that made no sense I'm sorry, like I said i'm still new to this. To me indefinite integrals don't really mean much. I grasp that an integral is the area under a line or curve (a definite integral), but it just seems indefinite integrals are the half-assed version of a definite integral. What else does it tell you?
 
  • #8
954
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are the two answers equal except for the constant +C? If so, how would I manipulate the constant to have them be identical?
Manipulating your "calculator answer":
[tex]\frac{x}{2}\, -\, \frac{1}{4} ln |2x+1|\, +\, C = \frac{x}{2}\, -\, \frac{1}{4} ln (|2||x + \frac{1}{2}|)\, +\, C = \frac{x}{2}\, -\, \frac{1}{4} ln |x + \frac{1}{2}| \,-\, \frac{1}{4} ln 2 \,+\, C = \frac{x}{2}\, -\, \frac{1}{4} ln |x + \frac{1}{2}|\, +\, C'[/tex]
 
  • #9
288
3
Manipulating your "calculator answer":
[tex]\frac{x}{2}\, -\, \frac{1}{4} ln |2x+1|\, +\, C = \frac{x}{2}\, -\, \frac{1}{4} ln (|2||x + \frac{1}{2}|)\, +\, C = \frac{x}{2}\, -\, \frac{1}{4} ln |x + \frac{1}{2}| \,-\, \frac{1}{4} ln 2 \,+\, C = \frac{x}{2}\, -\, \frac{1}{4} ln |x + \frac{1}{2}|\, +\, C'[/tex]
Thanks! god it seems to me most of the time that calculus is easy... it's the algebra that I took in 1996 that's what gets me (and most of my class actually).

But thanks again.
 
  • #10
288
3
when I take the derivative of either my answer or the calculator answer I get:

[tex]\frac{x}{2}[/tex] - [tex]\frac{1}{4x+2}[/tex]

But I can't seem to manipulate that answer to get the original fraction:

[tex]\frac{x}{2x+1}[/tex]


Any ideas?
 
  • #11
CompuChip
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You made a little mistake in your calculation (you copied the x/2 instead of differentiating it).

Then, how do you add two fractions (like 1/3 and 2/5)?
 
  • #12
288
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ah, got it, thanks!
 
  • #13
1,568
0
from that i had, 1/2 [tex]\int[/tex]dx - 1/2[tex]\int[/tex]dx/(x+1/2)

finally x/2 - 1/4 ln lx+1/2l + C


My calculator keeps telling me it should be x/2 - 1/4 ln l2x+1l + C and every time try using a different trick i get a different answer.
The "two" answers are in reality one and the same answer, with C-ln(2)=C'
 

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