Integral Homework Help: Stuck on t^4+16

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SUMMARY

The integral \(\int \frac{1}{t^4 + 16} \, dt\) presents significant challenges, particularly when attempting substitutions such as \(t^2 = 4u\) to simplify the expression. Users in the discussion highlighted the complexity of the integral, noting that it can lead to complicated results involving logarithmic and inverse tangent functions. The hint provided by dextercioby suggests factoring \(t^4 + 16\) into \((t^2 + 4)^2 - 8t^2\) and using partial fractions, but this approach also leads to complications, including imaginary numbers in the solution.

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Homework Statement


\int\frac{1}{t^4+16}\,dt

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
 
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Thanks dextercioby. If I understand your hint, I would factor (t^2+4)^2-8t^2 into two factors, use partial fractions, and then integrate the two expressions that would result?
 
That doesn't seem to work. I end up with imaginaries in the expression, unless I am doing something wrong.
 
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.
 
Ok, I think I've got it... Wow. This is so much more complicated than any other expression. Wow.
 
I'm not sure if that's what dex wanted.

But either way, you're doing something wrong.

\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}. Not very nice.
 

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