# Homework: Indefinite Integral

Evaluate the indefinite integral:

[(e^(4x))/(e^(8x))+9]dx

-I think that u=e^(2x)
so then du=e^(2x)dx
then the answer would end up being [(e^(4x)+9)/(-1)]^(-1)

but it was incorrect; I think that my u might be wrong and that's where the problem is, but I am not sure. Please help, thank you!

as in

$$\int (\frac{e^{4x}}{e^{8x}}+9)dx ?$$
$$\int (e^{4x-8x}+9)dx = - \frac{1}{4}e^{-4x}+9x$$

or as

$$\int (\frac{e^{4x}}{e^{8x}+9})dx ?$$
$$u=e^{8x}+9$$
$$du=8e^{8x}$$
and then i don't know how to solve this =D
If that's what you were asking, you got a tough one.
[still thinking]

It looks like an integration by parts question, or I am sleepy and can't see the answer D=
But integration by parts doesn't work in my case still...

Last edited:
correction on problem

the second part is the one that we need help on...thank you!!

$$u=e^{4x}$$
$$du=4e^{4x}$$

$$\frac{1}{4} \int(\frac{1}{u^{2}+3^{2}})du$$

Can you finish it from there?

Last edited:
$$u=e^{4x}$$
$$du=4e^{4x}$$

$$\frac{1}{4} \int(\frac{1}{u^{2}+3^{2}})du$$

Can you finish it from there?

As I was saying.. I must've been smoking something....

I always get confused when taking antiderivatives of fractions...how do I go about doing that?

I always get confused when taking antiderivatives of fractions...how do I go about doing that?

natural log?

So, I ended up with 1/4lnabs((e^4x)^(2))+9) and got it wrong, what am I doing incorrectly?

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