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Stuck on an integral

  • Thread starter BrettB
  • Start date
  • #1
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0

Homework Statement


[tex] \int\frac{1}{t^4+16}\,dt [/tex]


Homework Equations


Various integration formulas. I am pretty stuck.


The Attempt at a Solution



I have tried various substitutions, with no success. I tried t^2 = 4u, hoping to get it into the form 1/(u^2+1) so I could use the arctan formula, but then du/dt = t/2 and so I don't have an equivalent expression when I substitute.

I have tried various other substitutions with no better luck. If anyone has a suggestion, I would be so grateful. This is the only one from this set I haven't been able to solve.

Thanks!
Brett
 

Answers and Replies

  • #2
dextercioby
Science Advisor
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HINT: [tex] t^4 +16=t^4 +8t^2 +16 -8t^2 [/tex]
 
  • #3
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Thanks dextercioby. If I understand your hint, I would factor [tex] (t^2+4)^2-8t^2 [/tex] into two factors, use partial fractions, and then integrate the two expressions that would result?
 
  • #4
7
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That doesn't seem to work. I end up with imaginaries in the expression, unless I am doing something wrong.
 
  • #5
529
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I don't want to put you off- but the Integrator gives an answer that is quite complicated- involving 2 logs and 2 inverse tan functions. Could it be that you've misread the question?

I would rate this integral as tough.
 
  • #6
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Ok, I think I've got it... Wow. This is so much more complicated than any other expression. Wow.
 
  • #7
Gib Z
Homework Helper
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I'm not sure if thats what dex wanted.

But either way, you're doing something wrong.

[tex]\frac{1}{t^4+16}= \frac{\frac{-t}{16\sqrt{2}} - \frac{1}{8}}{t^2 + 2\sqrt{2} +4} - \frac{\frac{1}{8} - \frac{t}{16\sqrt{2}}}{-t^2+2\sqrt{2}-4}[/tex]. Not very nice.
 

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