1. ## integral

Thanks!

I have a similar problem, I am trying to solve the way you suggested:

$
\displaystyle\int \frac{e^{6 x}}{e^{12 x} + 9} dx
$

let U=e^6x

$
\displaystyle\int \frac{U}{U^2+9}
$

now, if I pull a 1/9 out of the integral

I can get
$
\displaystyle\frac{1}{9} \int\frac{U}{\frac{U^2}{9}+1}
$

Aghh, and there is ALMOST an arctan there, if it wasnt for that U in the numerator!

What do I do here? Am I not supposed to use this same method?

2. Originally Posted by Vamz
Thanks!

I have a similar problem, I am trying to solve the way you suggested:

$
\displaystyle\int \frac{e^{6 x}}{e^{12 x} + 9} dx
$

let U=e^6x

$
\displaystyle\int \frac{U}{U^2+9}
$

now, if I pull a 1/9 out of the integral

I can get
$
\displaystyle\frac{1}{9} \int\frac{U}{\frac{U^2}{9}+1}
$

Aghh, and there is ALMOST an arctan there, if it wasnt for that U in the numerator!

What do I do here? Am I not supposed to use this same method?
$e^{6x}=3\tan(\theta)$

I think this should work. I didn't try it.

3. Originally Posted by Vamz
Thanks!

I have a similar problem, I am trying to solve the way you suggested:

$
\displaystyle\int \frac{e^{6 x}}{e^{12 x} + 9} dx
$

let U=e^6x

$
\displaystyle\int \frac{U}{U^2+9}
$

now, if I pull a 1/9 out of the integral

I can get
$
\displaystyle\frac{1}{9} \int\frac{U}{\frac{U^2}{9}+1}
$

Aghh, and there is ALMOST an arctan there, if it wasnt for that U in the numerator!

What do I do here? Am I not supposed to use this same method?
It should not be

if $u=e^{6x} \implies du=6e^{6x}dx$ This gives the integtal

$\displaystyle \frac{1}{6}\int\frac{du}{u^2+9}$