Fastest way to integrate: dx/ (1 + x^4 )
6 years agokidding..
6 years agothis is the same as dx / ( 1 + (x^2)^2 )
6 years agolet u = x^2 , du = 2x dx ,
6 years agoI don't think that substitution would help.
6 years agoits a start
6 years agox^2=tan x naw...
6 years agodu/(2x) = dx
6 years agoand what do you propose to do with that x ?
6 years agowe can solve for x, we know that u = x^2, so x = sqrt u
6 years agoso I get integral 1 / [ u^1/2 * (1 + u^2) ] du
6 years agothats a known integral
6 years agoyou have to check your table of integrals
6 years agoI don't like tables, this should be derivable...
6 years agoAgreed.
6 years agodont forget the 1/2 integral 1/(1+x^4) = 1/2 integral 1 / [ u^1/2 * (1 + u^2) ] du
6 years agoWhat if I add subtract x^2 , then seperate it. Okay this may work.
6 years agono i dont think that would work
6 years agolooks very complicated http://www.wolframalpha.com/input/?i=int%28dx%2F+%281+%2B+x^4+%29+%29
6 years ago\[I = 1/2 \int\limits ( 1+ x^2)/(1+x^4) + (1-x^2)/(1+x^4)\]
6 years agomaybe complete the square?
6 years agoNow we just multiply divide by x^2
6 years agoyou added and subtracted x^2 on top?
6 years agoI didn't think wolfram would be promising... completing the square would give\[\int\frac{dx}{(x^2-1)^2+2x^2}\]which would not help I don't think
6 years agohere http://answers.yahoo.com/question/index?qid=20080721154424AAYOrmH
6 years agothere doesnt look like theres a fast way
6 years agoOh dear.
6 years agoso the trick is add and subtract 2x^2 on the bottom, which forces a perfect square
6 years agoi wouldnt have thought of this , :0
6 years agowell, it turns out completing the square can pay off... might as well do this with PF if it's gonna be that big of a hassle
6 years agothen you can use partial fractions
6 years ago, can you generally complete the square on any function, only for quadratics, and perhaps even functions . here we got lucky
6 years agoI did that above, but I didn't see where to go from there :/
6 years ago, in fact, this can lead to a general integral formula for dx / ( 1 + x^(2n) )
6 years ago, for n > 1
6 years agook you want to do it step by step ill make a trade with you, you help me with my simple 'work' problem. its just an integral
6 years agothe units stump me , im not so confident about it
6 years agooh cool, wikipedia already has the formula http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions
6 years agoscroll down to dx/ ( 1 + x^(2n) )
6 years agoso you want to go through the steps on yahoo?
6 years agoOkay. I think this works too. Similar though. \[(1-x^2)/(1+x^4) = (1/x^2 -1)/(x^2 + 1/x^2) \] Now use \[x - 1/x = t => dt = numerator\] \[t^2 = 1/x^2 + x^2 -2 => t^2 + 2 = denominator\]
6 years agoi know whats going on, the homework was dx/ (x^4 - 1) , and you made a typo. no teacher would give this assignment for baby math people
6 years agoThat helps Right?
6 years agomaybe, did you do it out ?
6 years agoor the teacher made a typo, this assignment is probably too difficult for calculus students
6 years agoYep. Look the integral now becomes: \[dt/( 1+t^2)\]
6 years agoNo. No mistake. This does work. Wait I'll write all the steps.
6 years agooh ok
6 years agoyou should get the answer [1/(2√2)] arctan (√2x +1) + [1/(2√2)] arctan (√2x - 1) + c
6 years agoso what did you do ?
6 years agoI didnt write the final few steps. But its easy after that.
6 years agotheres a mistake your first line is wrong
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6 years ago