Integration problem

  • Thread starter Melawrghk
  • Start date
  • #1
145
0

Homework Statement


[tex]\int \frac{2e^{2t}*dt}{e^{4t}+12e^{2t}+32}[/tex]

The Attempt at a Solution


I factored the denominator, getting (e2t+4)(e2t+8), from that I was able to express the original equation in terms of two fractions:
[tex]\int \frac{A}{e^{2t}+4}[/tex]+[tex]\frac{B}{e^{2t}+8}[/tex]

I then found A to be -2 and B=4.

But when I put those back into the original equations, I still have integrals that have an 'e' on the bottom and I don't know what to do next...

Any help is appreciated.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,223
31
Put u=e2t, and then split into partial fractions.
 
  • #3
145
0
More partial fractions? But I can't. u+4 doesn't factor...?
 
  • #4
rock.freak667
Homework Helper
6,223
31
More partial fractions? But I can't. u+4 doesn't factor...?

[tex]
\int \frac{2e^{2t}}{e^{4t}+12e^{2t}+32}dt
[/tex]

u=e^2t => du/2 =e^2t dt

[tex]\int \frac{2e^{2t}}{e^{4t}+12e^{2t}+32}dt \equiv \frac{1}{2} \int \frac{1}{u^2+12u+32} du[/tex]

Now do partial fractions.
 
  • #5
145
0
I already did. It's in my procedure...
 
  • #6
rock.freak667
Homework Helper
6,223
31
Then you'd simply have to integrate

[tex]\frac{1}{8} \int (\frac{1}{u+4} - \frac{1}{u+8})du[/tex]
 

Related Threads on Integration problem

  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
13
Views
1K
  • Last Post
Replies
1
Views
972
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
9
Views
1K
Top