OpenStudy (anonymous):

Fastest way to integrate: dx/ (1 + x^4 )

5 years ago
OpenStudy (anonymous):

kidding..

5 years ago
OpenStudy (perl):

this is the same as dx / ( 1 + (x^2)^2 )

5 years ago
OpenStudy (perl):

let u = x^2 , du = 2x dx ,

5 years ago
OpenStudy (anonymous):

I don't think that substitution would help.

5 years ago
OpenStudy (perl):

its a start

5 years ago
OpenStudy (turingtest):

x^2=tan x naw...

5 years ago
OpenStudy (perl):

du/(2x) = dx

5 years ago
OpenStudy (turingtest):

and what do you propose to do with that x ?

5 years ago
OpenStudy (perl):

we can solve for x, we know that u = x^2, so x = sqrt u

5 years ago
OpenStudy (perl):

so I get integral 1 / [ u^1/2 * (1 + u^2) ] du

5 years ago
OpenStudy (perl):

thats a known integral

5 years ago
OpenStudy (perl):

you have to check your table of integrals

5 years ago
OpenStudy (turingtest):

I don't like tables, this should be derivable...

5 years ago
OpenStudy (anonymous):

Agreed.

5 years ago
OpenStudy (perl):

dont forget the 1/2 integral 1/(1+x^4) = 1/2 integral 1 / [ u^1/2 * (1 + u^2) ] du

5 years ago
OpenStudy (anonymous):

What if I add subtract x^2 , then seperate it. Okay this may work.

5 years ago
OpenStudy (perl):

no i dont think that would work

5 years ago
OpenStudy (anonymous):

looks very complicated http://www.wolframalpha.com/input/?i=int%28dx%2F+%281+%2B+x^4+%29+%29

5 years ago
OpenStudy (anonymous):

\[I = 1/2 \int\limits ( 1+ x^2)/(1+x^4) + (1-x^2)/(1+x^4)\]

5 years ago
OpenStudy (perl):

maybe complete the square?

5 years ago
OpenStudy (anonymous):

Now we just multiply divide by x^2

5 years ago
OpenStudy (perl):

you added and subtracted x^2 on top?

5 years ago
OpenStudy (turingtest):

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

5 years ago
OpenStudy (perl):

here http://answers.yahoo.com/question/index?qid=20080721154424AAYOrmH

5 years ago
OpenStudy (perl):

there doesnt look like theres a fast way

5 years ago
OpenStudy (anonymous):

Oh dear.

5 years ago
OpenStudy (perl):

so the trick is add and subtract 2x^2 on the bottom, which forces a perfect square

5 years ago
OpenStudy (perl):

i wouldnt have thought of this , :0

5 years ago
OpenStudy (turingtest):

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

5 years ago
OpenStudy (perl):

then you can use partial fractions

5 years ago
OpenStudy (perl):

, can you generally complete the square on any function, only for quadratics, and perhaps even functions . here we got lucky

5 years ago
OpenStudy (turingtest):

I did that above, but I didn't see where to go from there :/

5 years ago
OpenStudy (perl):

, in fact, this can lead to a general integral formula for dx / ( 1 + x^(2n) )

5 years ago
OpenStudy (perl):

, for n > 1

5 years ago
OpenStudy (perl):

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

5 years ago
OpenStudy (perl):

the units stump me , im not so confident about it

5 years ago
OpenStudy (perl):

oh cool, wikipedia already has the formula http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions

5 years ago
OpenStudy (perl):

scroll down to dx/ ( 1 + x^(2n) )

5 years ago
OpenStudy (perl):

so you want to go through the steps on yahoo?

5 years ago
OpenStudy (anonymous):

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 \[x - 1/x = t => dt = numerator\] \[t^2 = 1/x^2 + x^2 -2 => t^2 + 2 = denominator\]

5 years ago
OpenStudy (perl):

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

5 years ago
OpenStudy (anonymous):

That helps Right?

5 years ago
OpenStudy (perl):

maybe, did you do it out ?

5 years ago
OpenStudy (perl):

or the teacher made a typo, this assignment is probably too difficult for calculus students

5 years ago
OpenStudy (anonymous):

Yep. Look the integral now becomes: \[dt/( 1+t^2)\]

5 years ago
OpenStudy (anonymous):

No. No mistake. This does work. Wait I'll write all the steps.

5 years ago
OpenStudy (perl):

oh ok

5 years ago
OpenStudy (perl):

you should get the answer [1/(2√2)] arctan (√2x +1) + [1/(2√2)] arctan (√2x - 1) + c

5 years ago
OpenStudy (perl):

so what did you do ?

5 years ago
OpenStudy (anonymous):
5 years ago

OpenStudy (anonymous):

I didnt write the final few steps. But its easy after that.

5 years ago
OpenStudy (perl):

theres a mistake your first line is wrong

5 years ago
OpenStudy (perl):

|dw:1345235836378:dw|

5 years ago