Integral of 1/(x^4+1). I think it is solved with partial fractions.

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Integral of 1/(x^4+1). I think it is solved with partial fractions.

Mathematics
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I don't think you can factor the denominator so we can't use partial fraction. Trig Substitution might work
x= tan(theta) dx= sec^2(theta) d-theta
is this a complex variables class or real?

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It is real. I got to this point: \[\int\limits\frac{\sec^{2}(\theta)}{(\tan^{4}(\theta)+1)}\; d\theta\]
yikes. i found the answer using maple and it is long and ugly. if you have to s how the steps try here http://answers.yahoo.com/question/index?qid=20080721154424AAYOrmH
someone is clearly out to get you with this problem. makes my head spin
Thank you for the link, it's in a take home part of a final in calculus 2 :(. I found it more simply in the end of this page: http://kr.cs.ait.ac.th/~radok/math/mat6/calc4.htm It says that even Leibnitz found this a troublesome problem :). They use partial fractions as such: \[x^4+1=(x^2+1)^2-2x^2=(x^2+1+\sqrt{2}x)(x^2+1-\sqrt{2}x)\] Can you explain how to get the third step?
yes i think. \[(x^2+1)^2-2x^2\] is the difference of two squares \[a^2-b^2=(a+b)(a-b)\] in this case \[a=x^2+1, b=\sqrt{2}x\]
thats should do it yes?
What about the 2 in 2x^2?
\[(\sqrt{2}x)^2=2x^2\] that is why \[b=\sqrt{2}x\]
Clever, that is why is \sqrt{2}
Thank you.
good luck!

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