Homework: Indefinite Integral

  • Thread starter MillerL7
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  • #1
MillerL7
14
0
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!
 

Answers and Replies

  • #2
Crazy Tosser
176
0
as in

[tex]\int (\frac{e^{4x}}{e^{8x}}+9)dx ?[/tex]
[tex]\int (e^{4x-8x}+9)dx = - \frac{1}{4}e^{-4x}+9x[/tex]

or as

[tex]\int (\frac{e^{4x}}{e^{8x}+9})dx ?[/tex]
[tex]u=e^{8x}+9[/tex]
[tex]du=8e^{8x}[/tex]
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:
  • #3
MillerL7
14
0
correction on problem

the second part is the one that we need help on...thank you!!
 
  • #4
exk
119
0
[tex]u=e^{4x}[/tex]
[tex]du=4e^{4x}[/tex]

[tex]
\frac{1}{4} \int(\frac{1}{u^{2}+3^{2}})du
[/tex]

Can you finish it from there?
 
Last edited:
  • #5
Crazy Tosser
176
0
[tex]u=e^{4x}[/tex]
[tex]du=4e^{4x}[/tex]

[tex]
\frac{1}{4} \int(\frac{1}{u^{2}+3^{2}})du
[/tex]

Can you finish it from there?

As I was saying.. I must've been smoking something....
 
  • #6
MillerL7
14
0
I always get confused when taking antiderivatives of fractions...how do I go about doing that?
 
  • #7
Crazy Tosser
176
0
I always get confused when taking antiderivatives of fractions...how do I go about doing that?

natural log?
 
  • #8
MillerL7
14
0
So, I ended up with 1/4lnabs((e^4x)^(2))+9) and got it wrong, what am I doing incorrectly?
 
  • #9
exk
119
0
[tex]\int\frac{1}{u^{2}+a^{2}}=\frac{1}{a}tan^{-1}(\frac{u}{a})+C[/tex]
 

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