Integration u-substitution Problem

  • Thread starter FallingMan
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In summary, it seems like the u in the denominator of the equation should not be there, but it is because e6x is equivalent to u2.
  • #1
FallingMan
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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. :)
 
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  • #3
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
[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
FallingMan said:
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
eumyang said:
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
FallingMan said:
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
eumyang said:
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
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
eumyang said:
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
FallingMan said:
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
gb7nash said:
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:

1. What is u-substitution in integration?

U-substitution, also known as the substitution method, is a technique used to simplify and solve integration problems. It involves replacing a complicated expression in the integrand with a single variable, typically denoted as u, in order to make the integration process easier.

2. When should I use u-substitution in integration?

U-substitution is best used when the integrand contains a function that can be rewritten in terms of a single variable. This is often the case when the integrand contains a composite function, such as a polynomial, trigonometric, or exponential function.

3. How do I choose the right u for u-substitution?

The key to choosing the right u for u-substitution is to look for a function within the integrand that has a derivative that is also present in the integrand. This allows for the use of the chain rule, making the integration process simpler.

4. Can I use u-substitution for definite integrals?

Yes, u-substitution can be used for both indefinite and definite integrals. When using u-substitution for a definite integral, be sure to change the limits of integration to match the new variable, u.

5. Are there any common mistakes to avoid when using u-substitution?

One common mistake to avoid when using u-substitution is forgetting to change the limits of integration when solving a definite integral. Another mistake is not properly applying the chain rule, resulting in an incorrect final answer. It is important to carefully follow the steps of u-substitution to avoid these errors.

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