## anonymous 5 years ago Integral of 1/(x^4+1). I think it is solved with partial fractions.

1. anonymous

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

2. anonymous

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

3. anonymous

is this a complex variables class or real?

4. anonymous

It is real. I got to this point: $\int\limits\frac{\sec^{2}(\theta)}{(\tan^{4}(\theta)+1)}\; d\theta$

5. anonymous

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

6. anonymous

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

7. anonymous

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?

8. anonymous

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

9. anonymous

thats should do it yes?

10. anonymous

What about the 2 in 2x^2?

11. anonymous

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

12. anonymous

Clever, that is why is \sqrt{2}

13. anonymous

Thank you.

14. anonymous

good luck!