## anonymous 3 years ago Integrate (1+x^2)/(1+x^4)dx

1. anonymous

$\large \int\limits\frac{1+x^{2}}{1+x^{4}} dx$

2. anonymous

ugh why couldnt it be 1 - x^4 :p lol

3. anonymous

If it was it would have defeated the purpose of this integral. (:

4. anonymous

exactly why im complaining :p

5. anonymous

No need to complain. Just think of an intuitive method. :)

6. blockcolder

Completing the square of the denominator then a substitution is what I have in mind, but I'm not completely sure.

7. anonymous

Not that easy.

8. anonymous

There are two methods.

9. anonymous

http://www.wolframalpha.com/input/?i=integrate+%5Cfrac%7B1%2Bx%5E%7B2%7D%7D%7B1%2Bx%5E%7B4%7D%7D+dx (look at "Show steps") Well, this looks... difficult. I wonder if there's an easier way to do this, than splitting it into two integrals.

10. anonymous

okay...hmm $u = \frac{1}{1+x^4} \rightarrow du = \frac{4x^3}{(1+x^4)^2}$ $dv = 1+ x^2 \rightarrow v = \tan^{-!} x$ LOL really isn't easy :P

11. blockcolder

Separation, perhaps? $\frac{1}{1+x^4}+\frac{x^2}{1+x^4}$ Might this work?

12. anonymous

Not really. Wolfram Alpha method is tedious.

13. anonymous

You can try if it works; but I dont think so.

14. anonymous

the first term would be integrable i think

15. anonymous

$\large\frac{1+\frac1{x^2}}{x^2 + \frac{1}{x^2}} = \frac{1 + \frac1{x^2}}{\left(x -\frac{1}x\right)+2}$This should do it.

16. blockcolder

Wow. O_O

17. anonymous

$\large\frac{1+\frac1{x^2}}{x^2 + \frac{1}{x^2}} = \frac{1 + \frac1{x^2}}{\left(x -\frac{1}x\right)^2+2}$

18. anonymous

Very nice Ishaan94; trig sub would work as well (: But that is easier.