Integration u-substitution Problem

[FallingMan]

Homework Statement

\int \frac{e^6^x}{e^1^2^x+64}dx

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. :)

 

[LCKurtz]

Try u = e^6x.

 

[FallingMan]

Okay, let's see...

If u=e^6^x then

du=6e^6^xdx.

So,

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

I'm still a bit confused...

 

[gb7nash]

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

= \frac{1}{6}\int \frac{u}{u^2+64}\times\frac{du}{u}

= ...

 

[eumyang]

Okay, let's see...

If u=e^6^x then

du=6e^6^xdx.

So,

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

I'm still a bit confused...

That e^6x in the denominator should not be there, should it?

[STRIKE]If u = e^6x, what is e^12x in terms of u?
[/STRIKE]

EDIT: Never mind. gb7nash gave it away.

 

[FallingMan]

That e^6x in the denominator should not be there.

If u = e^6x, what is e^12x in terms of u?

Err, shouldn't it be u²? (I think Nash was getting at something similar).. also, why shouldn't the e^6x 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?

 

[eumyang]

Err, shouldn't it be u²? (I think Nash was getting at something similar).. also, why shouldn't the e^6x 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.

\int \frac{e^6^x}{e^1^2^x+64}dx

Let u = e^6x.
Then du = 6e^6x dx.
So
\frac{du}{6} = e^{6x} dx.

Which means the integral becomes
\frac{1}{6} \int \frac{du}{u^2+64}EDIT: Fixed.

 

[FallingMan]

Yes, that's what I mean.

\int \frac{e^6^x}{e^1^2^x+64}dx

Let u = e^6x.
Then du = 6e^6x dx.
So
\frac{du}{6} = e^{6x} dx.

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


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.

 

[eumyang]

If you are studying from a calculus textbook, look for a table of integrals. There is one that starts like this:
\int \frac{du}{u^2 + a^2} = ...
(or something similar).

 

[FallingMan]

If you are studying from a calculus textbook, look for a table of integrals. There is one that starts like this:
\int \frac{du}{u^2 + a^2} = ...
(or something similar).

Oh... like...

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

Like that?

 

[gb7nash]

Oh... like...

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

Like that?

Looks right. Just remember to convert back to x.

 

[FallingMan]

Looks right. Just remember to convert back to x.

=\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

That seems right.. thank you for the help guys!