Question

Integrate (1+x^2)/(1+x^4)dx

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

exactly why im complaining :p

Answer 5

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

Answer 6

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

Answer 7

Not that easy.

Answer 8

There are two methods.

Answer 9

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.

Answer 10

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

Answer 11

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

Answer 12

Not really.
Wolfram Alpha method is tedious.

Answer 13

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

Answer 14

the first term would be integrable i think

Answer 15

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

Answer 16

Wow. O_O

Answer 17

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

Answer 18

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