# Stuck on an integral

## 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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dextercioby
Homework Helper
HINT: $$t^4 +16=t^4 +8t^2 +16 -8t^2$$

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.

Gib Z
Homework Helper
I'm not sure if thats 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.