# Math Help - Integration... Ruins my life...

1. ## Integration... Ruins my life...

I missed a week of lectures cuz I messed up my shoulder now I'm struggling to do my assignment...

$
\int \frac{e^{4 x}}{e^{8 x} + 36} dx
$

Can anyone please explain how to do that in details?
I figure you use substitution but it doesn't work out and I cant find another example similar to that in my text or posted notes.

2. Originally Posted by Lolcats
$
\int \frac{e^{4 x}}{e^{8 x} + 36} dx
$
This is clasic $\arctan(u)$ form.

3. Open the spoiler, if and only if you need more help.

Spoiler:
$e^{8x}=\left( e^{4x} \right)^{2},$ and substitute $6t=e^{4x}.$

4. See thats the thing, I dont know what a classic arctan thing is? I've missed my entire week of lectures so I'm pretty much in the dark to how to do it. Arctan i know the formula and I can follow the steps but I don't know the basis behind it. As to (e^4x)^2 I've realized that but why replace it with 6t? Why not just replace it with u then use du=dx?

Sorry I know its a struggle to explain a concept over a forum, I just dont have many other resources, just I spent 2 hours at a tutorial in which the TA decided to spend over an hour on one person, being totally oblivious to the other people in the room with questions.

5. Find the derivative of $\frac{1}
{{24}}\arctan \left( {\frac{{e^{4x} }}
{6}} \right)$

What do you see?

If you cannot do this derivative then this problem is not for you.

6. uhmmm k I get

(e^(4x))/(1+((e^(8x))/36)36...

I don't see how that gets me any closer, regardless if I couldn't do that I wouldn't have much of a choice cuz its on my assignment...

7. k this is what im getting...
$
\int \displaystyle{\frac{u}{u^{2}+36}\frac{du}{4e^{4x}} }
$

..then im stuck

u=e^(4x) and du=4e^(4x)

8. Originally Posted by Lolcats
k this is what im getting...
$
\int \displaystyle{\frac{u}{u^{2}+36}\frac{du}{4e^{4x}} }
$

..then im stuck

u=e^(4x) and du=4e^(4x)
Replace that $e^{4x}$ in the denominator with u!