Question

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

Answer 1

I don't think you can factor the denominator so we can't use partial fraction. Trig Substitution might work

Answer 2

x= tan(theta)
dx= sec^2(theta) d-theta

Answer 3

is this a complex variables class or real?

Answer 4

It is real. I got to this point:

\[\int\limits\frac{\sec^{2}(\theta)}{(\tan^{4}(\theta)+1)}\; d\theta\]

Answer 5

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

Answer 6

someone is clearly out to get you with this problem. makes my head spin

Answer 7

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?

Answer 8

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 \[a=x^2+1, b=\sqrt{2}x\]

Answer 9

thats should do it yes?

Answer 10

What about the 2 in 2x^2?

Answer 11

\[(\sqrt{2}x)^2=2x^2\] that is why \[b=\sqrt{2}x\]

Answer 12

Clever, that is why is \sqrt{2}

Answer 13

Thank you.

Answer 14

good luck!