Integration u-substitution Problem

  • Thread starter FallingMan
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  • #1
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Homework Statement




[tex]\int \frac{e^6^x}{e^1^2^x+64}dx[/tex]


The Attempt at a Solution



Well, it seems like it might be a u-substitution problem, but I think I have to rewrite it somehow to get the numerator to cancel? If anyone can give me a hint on what to choose for u, or maybe how to split it up, it would be great. :)
 

Answers and Replies

  • #3
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Okay, let's see...

If [tex]u=e^6^x[/tex] then

[tex]du=6e^6^xdx[/tex].

So,

[tex]\frac{1}{6}\int \frac{u}{e^1^2^x+64}\times\frac{du}{e^6^x}[/tex]

I'm still a bit confused...
 
  • #4
gb7nash
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[tex]\frac{1}{6}\int \frac{u}{e^1^2^x+64}\times\frac{du}{e^6^x}[/tex]

= [tex]\frac{1}{6}\int \frac{u}{u^2+64}\times\frac{du}{u}[/tex]

= ...
 
  • #5
eumyang
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Okay, let's see...

If [tex]u=e^6^x[/tex] then

[tex]du=6e^6^xdx[/tex].

So,

[tex]\frac{1}{6}\int \frac{u}{e^1^2^x+64}\times\frac{du}{e^6^x}[/tex]

I'm still a bit confused...
That e6x in the denominator should not be there, should it?

[STRIKE]If u = e6x, what is e12x in terms of u?
[/STRIKE]

EDIT: Never mind. gb7nash gave it away.
 
  • #6
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That e6x in the denominator should not be there.

If u = e6x, what is e12x in terms of u?
Err, shouldn't it be u2? (I think Nash was getting at something similar).. also, why shouldn't the e6x be in the denominator? I was solving for dx so - then you can get rid of the dx term in the integral, right?

EDIT: Oh, you mean it shouldn't be there because you can replace it with a u?
 
  • #7
eumyang
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Err, shouldn't it be u2? (I think Nash was getting at something similar).. also, why shouldn't the e6x be in the denominator? I was solving for dx so - then you can get rid of the dx term in the integral, right?

EDIT: Oh, you mean it shouldn't be there because you can replace it with a u?
Yes, that's what I mean.

[tex]\int \frac{e^6^x}{e^1^2^x+64}dx[/tex]

Let u = e6x.
Then du = 6e6x dx.
So
[tex]\frac{du}{6} = e^{6x} dx[/tex].

Which means the integral becomes
[tex]\frac{1}{6} \int \frac{du}{u^2+64}[/tex]


EDIT: Fixed.
 
  • #8
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Yes, that's what I mean.

[tex]\int \frac{e^6^x}{e^1^2^x+64}dx[/tex]

Let u = e6x.
Then du = 6e6x dx.
So
[tex]\frac{du}{6} = e^{6x} dx[/tex].

Which means the integral becomes
[tex]\frac{1}{6} \int \frac{du}{u^2+64}[/tex]


EDIT: Fixed.
Hmmm, interesting. Is there some rule that I am not aware of for dealing with this situation. Wolfram-alpha is giving me some way to integrate it that I don't really understand... At least I understand how to get to this point.

I might need some more time to figure out the rest.
 
  • #9
eumyang
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If you are studying from a calculus textbook, look for a table of integrals. There is one that starts like this:
[tex]\int \frac{du}{u^2 + a^2} = ...[/tex]
(or something similar).
 
  • #10
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If you are studying from a calculus textbook, look for a table of integrals. There is one that starts like this:
[tex]\int \frac{du}{u^2 + a^2} = ...[/tex]
(or something similar).
Oh.... like...

[tex]\int \frac{du}{u^2 + a^2} = \frac{tan^-^1(\frac{u}{a})}{a}+C[/tex]

Like that?
 
  • #11
gb7nash
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Oh.... like...

[tex]\int \frac{du}{u^2 + a^2} = \frac{tan^-^1(\frac{u}{a})}{a}+C[/tex]

Like that?
Looks right. Just remember to convert back to x.
 
  • #12
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Looks right. Just remember to convert back to x.
[tex]=\frac{1}{6} \int\frac{1}{u^2+8^2}du = \frac{1}{6}\times\frac{tan^-^1\frac{u}{8}}{8}=\frac{1}{6}\times\frac{tan^-^1(\frac{e^6^x}{8})}{8}+C[/tex]

That seems right.. thank you for the help guys!
 
Last edited:

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