What is the Indefinite Integral of [(e^(4x))/(e^(8x))+9]dx?

[MillerL7]

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!

 

[Crazy Tosser]

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...

 

[MillerL7]

correction on problem

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

 

[exk]

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

<br /> \frac{1}{4} \int(\frac{1}{u^{2}+3^{2}})du <br />

Can you finish it from there?

 

[Crazy Tosser]

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

<br /> \frac{1}{4} \int(\frac{1}{u^{2}+3^{2}})du <br />

Can you finish it from there?

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

 

[MillerL7]

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

 

[Crazy Tosser]

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

natural log?

 

[MillerL7]

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

 

[exk]

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