# Integral Homework Help: Stuck on t^4+16

• BrettB
In summary, the integral \int\frac{1}{t^4+16}\,dt is a difficult problem that requires various substitutions and the use of partial fractions. The solution involves a complicated expression with logarithms and inverse tangent functions.

## 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

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.

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.