# Integration u-substitution Problem

#### FallingMan

1. Homework Statement

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

3. 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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Homework Helper
Gold Member
Try u = e6x.

#### 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

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

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

= ...

#### eumyang

Homework Helper
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 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.

#### FallingMan

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?

#### eumyang

Homework Helper
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.

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

Let u = e6x.
Then du = 6e6x 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 = e6x.
Then du = 6e6x 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

Homework Helper
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

Homework Helper
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!

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