Integration, u substitution, 1/u

  • Thread starter Gibybo
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  • #1
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[SOLVED] Integration, u substitution, 1/u

-- +C at the end of the integral solutions, I can't seem to add it in the LaTeX thing --

Homework Statement


#1 [tex]\int\frac{1}{8-4x}dx[/tex]
#2 [tex]\int\frac{1}{2x}dx[/tex]

The Attempt at a Solution


#1
Rewrite algebraically:
[tex]\int\frac{1}{x-2}*\frac{-1}{4}dx[/tex]

Pull out constant:
[tex]\frac{-1}{4}\int\frac{1}{x-2}dx[/tex]

Usub, u=x-2, du=1*dx,
[tex]\frac{-1}{4}\int\frac{1}{u}du[/tex]

[tex]\frac{-ln(u)}{4}[/tex] -> Answer: [tex]\frac{-ln(x-2)}{4}[/tex]

#2
Pull out constant: [tex]\frac{1}{2}*\int\frac{1}{x}dx[/tex]

Answer: [tex]\frac{ln(x)}{2}[/tex]

--OR, if I don't pull out constants--

#1
Usub, u=8-4x, du=-4dx,
[tex]\frac{-1}{4}*\int\frac{1}{u}du[/tex]

[tex]\frac{-ln(u)}{4}[/tex] -> Answer: [tex]\frac{-ln(8-4x)}{4}[/tex], which =/= [tex]\frac{-ln(x-2)}{4}[/tex]

#2
Usub, u=2x, du = 2dx,
[tex]\frac{1}{2}[/tex][tex]\int\frac{1}{u}du[/tex]

[tex]\frac{ln(u)}{2}[/tex] -> Answer: [tex]\frac{ln(2x)}{2}[/tex], which =/= [tex]\frac{ln(x)}{2}[/tex]

I know the answers are -1/4*ln(8-4x) and ln(x)/2 respectively, but I don't understand why I cannot pull out the constant for the first one, yet I am required to pull it out for the second?

Bonus question: Why does my TI-89 integrate #1 incorrectly, but #2 correctly? :(
 
Last edited:

Answers and Replies

  • #2
1,752
1
(8-4x),x ... did you mean dx? cuz i'm confused.

#2 is correct for sure, let me go back to # 1.

i'm getting dizzy reading your text but from what i see, your algebra-manipulation should have been for #1

[tex]\frac{1}{4}\int\frac{1}{2-x}[/tex]
 
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  • #3
13
0
Yeah, by integrate(4x,x) I would mean 4x*dx. It's standard notation when typing integrals for calculators/CAS's etc, I did it that way because I was thinking the forums might automatically convert it to the proper mathematical symbols.
 
  • #4
1,752
1
Yeah, by integrate(4x,x) I would mean 4x*dx. It's standard notation when typing integrals for calculators/CAS's etc, I did it that way because I was thinking the forums might automatically convert it to the proper mathematical symbols.
oh sorry, i did not know that! :)
 
  • #5
13
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i'm getting dizzy reading your text but from what i see, your algebra-manipulation should have been for #1

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

Ah hah, that's how you make it pretty!

Anyway yeah, I have [tex]\frac{-1}{4}\int\frac{1}{x-2}[/tex]
which is equivalent and still valid, right?
 
  • #6
1,752
1
Ah hah, that's how you make it pretty!

Anyway yeah, I have [tex]\frac{-1}{4}\int\frac{1}{x-2}[/tex]
which is equivalent and still valid, right?
Introducing LaTeX Math Typesetting - https://www.physicsforums.com/showthread.php?t=8997 it gets addicting!!!

yep, don't forget + C ;) .. -1 or maybe the whole problem will be counted wrong, lol :D
 
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  • #7
492
1
when you do your final integrations the ln shouldn't be in the denominator of the fraction.

It's (-1/4)*(lnx-2)+c which gives -ln(x-2)/4 + C the same goes for the second one.
 
  • #8
13
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when you do your final integrations the ln shouldn't be in the denominator of the fraction.

It's (-1/4)*(lnx-2)+c which gives -ln(x-2)/4 + C the same goes for the second one.

Oh sorry, just an error while converting to [tex] images I think.

-ln(x-2)/4+c is incorrect though, which is my problem.
 
  • #9
329
1
I'm not sure why you pulled the constant out in the first place. I would go u = 8-4x du= -4dx and then write (-1/4)*integral 1/u du.

Integrate and get (-1/4)ln(8-4x)

the second one I would make u = 2x du = 2dx and then write the integral as 1/2 * integral 1/u * du and the integrate and get (1/2) ln (2x)
 
  • #10
13
0
I'm not sure why you pulled the constant out in the first place. I would go u = 8-4x du= -4dx and then write (-1/4)*integral 1/u du.

Integrate and get (-1/4)ln(8-4x)

the second one I would make u = 2x du = 2dx and then write the integral as 1/2 * integral 1/u * du and the integrate and get (1/2) ln (2x)

Right, I did both of them both ways. If you do not pull out the constants, you get the correct answer for #1. However, #2 is not (1/2)*ln(2x), it is ln(x)/2.

Summary:
Pull out constants:
#1 wrong, #2 right

Do NOT pull out constants:
#1 right, #2 wrong
 
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  • #11
1,752
1
there really is no benefit in simplifying this problem, i would just go at what you're given.
 
  • #12
13
0
there really is no benefit in simplifying this problem, i would just go at what you're given.

I'm not sure what you're saying. I shouldn't pull out constants?

Why is it wrong to pull them out then?
And how do you correctly do #2 without pulling out constants first?
 
  • #13
1,752
1
for 1, integrate by u subst., but 2 you have to pull out the constant.

i just don't see a point in pulling out a constant for 1 b/c you run the risk of making algebraic mistakes, that's all really.
 
  • #14
13
0
for 1, integrate by u subst., but 2 you have to pull out the constant.

i just don't see a point in pulling out a constant for 1 b/c you run the risk of making algebraic mistakes, that's all really.

I agree, but why is it wrong when I do? I think we can agree that there are no algebraic mistakes there. Why do I HAVE to pull them out for #2?
 
  • #15
Avodyne
Science Advisor
1,396
88
ln(2x) = ln(x) + ln(2) = ln(x) + C, with C = ln(2). Since indefinite integrals are only defined up to an additive constant, your two answers are the same. This is true for both problems.
 
  • #16
13
0
ln(2x) = ln(x) + ln(2) = ln(x) + C, with C = ln(2). Since indefinite integrals are only defined up to an additive constant, your two answers are the same. This is true for both problems.

Ah hah! Thanks :) [hits self on head]
 
  • #17
Avodyne
Science Advisor
1,396
88
You're welcome! :) (It's always the little things that cause trouble ...)
 

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