Here's the question you clicked on:
Omniscience
Integrate (1+x^2)/(1+x^4)dx
\[\large \int\limits\frac{1+x^{2}}{1+x^{4}} dx\]
ugh why couldnt it be 1 - x^4 :p lol
If it was it would have defeated the purpose of this integral. (:
exactly why im complaining :p
No need to complain. Just think of an intuitive method. :)
Completing the square of the denominator then a substitution is what I have in mind, but I'm not completely sure.
There are two methods.
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.
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
Separation, perhaps? \[\frac{1}{1+x^4}+\frac{x^2}{1+x^4}\] Might this work?
Not really. Wolfram Alpha method is tedious.
You can try if it works; but I dont think so.
the first term would be integrable i think
\[\large\frac{1+\frac1{x^2}}{x^2 + \frac{1}{x^2}} = \frac{1 + \frac1{x^2}}{\left(x -\frac{1}x\right)+2} \]This should do it.
\[\large\frac{1+\frac1{x^2}}{x^2 + \frac{1}{x^2}} = \frac{1 + \frac1{x^2}}{\left(x -\frac{1}x\right)^2+2}\]
Very nice Ishaan94; trig sub would work as well (: But that is easier.