Question

Fastest way to integrate:

dx/ (1 + x^4 )

Answer 1

kidding..

Answer 2

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

Answer 3

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

Answer 4

I don't think that substitution would help.

Answer 5

its a start

Answer 6

x^2=tan x
naw...

Answer 7

du/(2x) = dx

Answer 8

and what do you propose to do with that x ?

Answer 9

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

Answer 10

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

Answer 11

thats a known integral

Answer 12

you have to check your table of integrals

Answer 13

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

Answer 14

Agreed.

Answer 15

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

Answer 16

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

Answer 17

no i dont think that would work

Answer 18

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

Answer 19

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

Answer 20

maybe complete the square?

Answer 21

Now we just multiply divide by x^2

Answer 22

you added and subtracted x^2 on top?

Answer 23

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

Answer 24

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

Answer 25

there doesnt look like theres a fast way

Answer 26

Oh dear.

Answer 27

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

Answer 28

i wouldnt have thought of this , :0

Answer 29

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

Answer 30

then you can use partial fractions

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

, for n > 1

Answer 35

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

Answer 36

the units stump me , im not so confident about it

Answer 37

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

Answer 38

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

Answer 39

so you want to go through the steps on yahoo?

Answer 40

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

Answer 41

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

Answer 42

That helps Right?

Answer 43

maybe, did you do it out ?

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

oh ok

Answer 48

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

Answer 49

so what did you do ?

Answer 50

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

Answer 51

theres a mistake
your first line is wrong