Fastest way to integrate: dx/ (1 + x^4 )
kidding..
this is the same as dx / ( 1 + (x^2)^2 )
let u = x^2 , du = 2x dx ,
I don't think that substitution would help.
its a start
x^2=tan x naw...
du/(2x) = dx
and what do you propose to do with that x ?
we can solve for x, we know that u = x^2, so x = sqrt u
so I get integral 1 / [ u^1/2 * (1 + u^2) ] du
thats a known integral
you have to check your table of integrals
I don't like tables, this should be derivable...
Agreed.
dont forget the 1/2 integral 1/(1+x^4) = 1/2 integral 1 / [ u^1/2 * (1 + u^2) ] du
What if I add subtract x^2 , then seperate it. Okay this may work.
no i dont think that would work
looks very complicated http://www.wolframalpha.com/input/?i=int%28dx%2F+%281+%2B+x^4+%29+%29
\[I = 1/2 \int\limits ( 1+ x^2)/(1+x^4) + (1-x^2)/(1+x^4)\]
maybe complete the square?
Now we just multiply divide by x^2
you added and subtracted x^2 on top?
I 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
there doesnt look like theres a fast way
Oh dear.
so the trick is add and subtract 2x^2 on the bottom, which forces a perfect square
i wouldnt have thought of this , :0
well, 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
then you can use partial fractions
, can you generally complete the square on any function, only for quadratics, and perhaps even functions . here we got lucky
I did that above, but I didn't see where to go from there :/
, in fact, this can lead to a general integral formula for dx / ( 1 + x^(2n) )
, for n > 1
ok 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
the units stump me , im not so confident about it
oh cool, wikipedia already has the formula http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions
scroll down to dx/ ( 1 + x^(2n) )
so you want to go through the steps on yahoo?
Okay. I think this works too. Similar though. \[(1-x^2)/(1+x^4) = (1/x^2 -1)/(x^2 + 1/x^2) \] Now use \[x - 1/x = t => dt = numerator\] \[t^2 = 1/x^2 + x^2 -2 => t^2 + 2 = denominator\]
i 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
That helps Right?
maybe, did you do it out ?
or the teacher made a typo, this assignment is probably too difficult for calculus students
Yep. Look the integral now becomes: \[dt/( 1+t^2)\]
No. No mistake. This does work. Wait I'll write all the steps.
oh ok
you should get the answer [1/(2√2)] arctan (√2x +1) + [1/(2√2)] arctan (√2x - 1) + c
so what did you do ?
I didnt write the final few steps. But its easy after that.
theres a mistake your first line is wrong
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