anonymous
  • anonymous
Hi! Integrate 2x/(x^8+1) dx
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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primeralph
  • primeralph
I'll be honest, I can solve that......the solution is not neat......so don't expect anyone here to be able to finish it....it's long...........try wolfram....... but you still can try here though
anonymous
  • anonymous
can you tell me how to at least start it?
anonymous
  • anonymous
I feel like this will end up being an ln function, but I think there's a lot more to it than just that I'm sure. Let Ralph help you.

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anonymous
  • anonymous
does anyone know of a completed example of a problem like this I can go online and review or anything?
primeralph
  • primeralph
|dw:1368859626389:dw|
primeralph
  • primeralph
That's how you start.
anonymous
  • anonymous
can you tell me what the J is for? I have never ever seen that used before
primeralph
  • primeralph
|dw:1368859717310:dw|
primeralph
  • primeralph
J is just a multiplier. In this case, J = 1 , so you can remove it
anonymous
  • anonymous
(x^8+1) means: X^9
anonymous
  • anonymous
I highly doubt that it would be that simple felavin.
anonymous
  • anonymous
WHY? SMOKEYDABEAR.
anonymous
  • anonymous
could the next step be to let u=tan(theta) then du=sec^2(theta) d(theta)
primeralph
  • primeralph
@SinginDaCalc2Blues same thing.........you're looping
anonymous
  • anonymous
Yeah all that would do is put you right back to where you started
primeralph
  • primeralph
^^^ that girl is just too smart..........
anonymous
  • anonymous
Too stupid*
anonymous
  • anonymous
I'm just here to learn how to do it myself
anonymous
  • anonymous
are you people calling me stupid? seriously?
anonymous
  • anonymous
No, I was calling myself stupid xD
primeralph
  • primeralph
no, I didn't mean that.
primeralph
  • primeralph
okay, let's figure out an easier way to solve this problem.
primeralph
  • primeralph
@SinginDaCalc2Blues can you handle recursive integrals?
anonymous
  • anonymous
okay...well I appreciate the help from everyone but it's hard to understand the context of non-verbal communication sometimes...I was hoping you guys weren't calling me stupid...anyway...
primeralph
  • primeralph
@SinginDaCalc2Blues we really would never do that.
anonymous
  • anonymous
i have never heard the term recursive integral...while I probably know what your talking about, I don't know it in those terms...can you describe?
primeralph
  • primeralph
wait............a minute...why not just polar?
anonymous
  • anonymous
Polar coordinates? Actually that makes some sense...
anonymous
  • anonymous
are you saying turn the integral into a polar expression or something?
primeralph
  • primeralph
@smokeydabear tricky thing is, it returns the integrand to the form we had before....because using cos, cos = 1/sec.....
anonymous
  • anonymous
my professor is devious!
anonymous
  • anonymous
Your professor is pretty evil for giving you something like this
anonymous
  • anonymous
would u=x^2 du=2x dx indefinite integral of 1/(u^4+1) help? (sorry, my latex isn't working)
primeralph
  • primeralph
@SinginDaCalc2Blues I reduced it to......can you solve this?|dw:1368860855998:dw|
anonymous
  • anonymous
the int, called complex in
anonymous
  • anonymous
uhm if k is a constant, what is J again?
primeralph
  • primeralph
J is the variable of integration, just like U
primeralph
  • primeralph
U in U substitution
anonymous
  • anonymous
I just looked it up on wolfram... what is this monstrosity...
primeralph
  • primeralph
@smokeydabear I told you...........
anonymous
  • anonymous
How do you know where to even start with something like this? That's always been my problem
anonymous
  • anonymous
a good place to start is with Vodka :o)
primeralph
  • primeralph
@SinginDaCalc2Blues I asked you if you knew recursives because as you can see here, the trigs are repeating..which means recursion
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anonymous
  • anonymous
no, not vodka, start with despair.
anonymous
  • anonymous
in 30 minutes i will no longer be able to do integration but I will be able to slurrrr Russian! :o)~
anonymous
  • anonymous
vodka inhibits your ability to do math.
anonymous
  • anonymous
despair only temporarily inhibits your ability to do everything
primeralph
  • primeralph
@smokeydabear recursion works to start. for example|dw:1368861204284:dw|
anonymous
  • anonymous
Okay, so how would we use that to help us? I'm still confused :s
anonymous
  • anonymous
crazy problem o:
anonymous
  • anonymous
well long only :P
primeralph
  • primeralph
@smokeydabear it saves time because once you know that it will repeat, you can generalize..........
anonymous
  • anonymous
hey guys...i have an idea...gimme a sec...
primeralph
  • primeralph
.........meaning............|dw:1368861457508:dw|
stamp
  • stamp
u = x^2, du = 2x dx\[\int \frac{du}{u^4+1}\]
primeralph
  • primeralph
so, because they occur in trig functions, we can use it for the tan theta
primeralph
  • primeralph
@stamp did that
anonymous
  • anonymous
Ohhhhhh I see what you did with the sin's and the pi's @primeralph very clever! =D
anonymous
  • anonymous
now we just need to learn to use it effectively...
primeralph
  • primeralph
|dw:1368861979382:dw|
anonymous
  • anonymous
I gotcha Ralph, I noticed it too but I didn't want to be mean =P
anonymous
  • anonymous
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anonymous
  • anonymous
Does this little bit of trickyness open any doors?
anonymous
  • anonymous
yeah what stamp wrote I believe is wrong...it wouldn't be x^4...it would be (x^4)^2
anonymous
  • anonymous
Well what stamp wrote isn't technically wrong, but it isn't the method to solving this.
anonymous
  • anonymous
I basically factored out of the denominator an x^2...then did u-sub
anonymous
  • anonymous
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anonymous
  • anonymous
hi, the integrate is simple: Who can solve this integrate:\[\int\limits_{}^{} \frac{\tan Z }{ Z- \frac{ P }{ 4}}\] GAME IS OVER ..........
anonymous
  • anonymous
hello
anonymous
  • anonymous
hi...i'm only in calc 2...i don't know what Cauchy-Reimann equations are... I did however get my integral down to 1/(u^3+1)
anonymous
  • anonymous
In case you love twisting with partial fractions: \[\frac{ 2x }{ x^8 +1}=\frac{ 2x }{( x^4-\sqrt{2}x^2+1)( x^4+\sqrt{2}x^2+1)}\]
anonymous
  • anonymous
hi drawer...i think your way is looking promising...i think i was headed toward that...can you look at my .bmp and tell me how to start partial fraction with what i ended with?
anonymous
  • anonymous
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anonymous
  • anonymous
Then \[u^3+1=(u+1)(u^2-u+1)\], use partial fractions.
anonymous
  • anonymous
what do i do with the numerator?
anonymous
  • anonymous
\[\frac{ 1 }{ u^3+1 }=\frac{ A }{ u+1 }+\frac{ Bu+C }{ u^2-u+1 }\] =>\[A(u^2-u+1)+(Bu+C)(u+1)=1\] Then equate the coefficients and you'll get a system of linear equations in terms of A,B, and C. Solve it!
anonymous
  • anonymous
so then...after I do that, will i have technically several different integrals which will be easier to integrate?
anonymous
  • anonymous
Yes, you'll see that the degree of the polynomial of the numerator is always 1 less than that of the denominator, which must have sth to do with the natural logarithm (ln) function.
shubhamsrg
  • shubhamsrg
i havent read much what is written above, its too long, has the question been solved ?
anonymous
  • anonymous
ok drawar...the answer definitely has some nasty ln's...i will keep plugging away at it...one more question...
anonymous
  • anonymous
once i figure out what A, B, AND C equate to, don't I just plug those into the numerators and then treat one as a separate integral to solve?
shubhamsrg
  • shubhamsrg
has anyone tried x^2 =t ?
anonymous
  • anonymous
yeah shub...wayyy up at the top...we are past that now...but thanks
anonymous
  • anonymous
Yes @SinginDaCalc2Blues
shubhamsrg
  • shubhamsrg
well, why you didnt go forward with that method, after that substitution, we'll have dt/(t^4 +1) note that t^4 + 1 = (t^2 + i) (t^2 -i) also note that (t^2 +i) - ( t^ -i) = 2i , a constant
anonymous
  • anonymous
ok...i will keep plugging away and see if I can get it...this problem has been so crazy! :o)
anonymous
  • anonymous
are you using imaginary numbers shub?
anonymous
  • anonymous
is it a definite/indefinite integral in your original question?
shubhamsrg
  • shubhamsrg
yes, the very same. i can be treated as just another constant
anonymous
  • anonymous
indefinite
anonymous
  • anonymous
i have no idea how to use imaginary numbers...i'm only in calc 2..we haven't used those
shubhamsrg
  • shubhamsrg
I see, well use of iota will lead you to the solution withing 2 more steps, and in the end you can always replace the complex part with hyperbolic functions, or natural logarithms . But anyways, since you havent learnt about that, go with @drawar 's method (:
anonymous
  • anonymous
I wish I had learned about them...i just hope i can solve this
anonymous
  • anonymous
\[\int\frac{2x}{x^8+1}dx\]\[=\int \frac{1}{x^8+1}d(x^2)\]\[=\int \frac{1}{y^4+1}dy\]\[=\frac{1}{2}\int \frac{y^2+1-y^2 +1}{y^4+1} dy\]\[=\frac{1}{2}(\int \frac{y^2+1}{y^4+1}dy-\int\frac{y^2 -1}{y^4+1} dy)\]\[=\frac{1}{2}(\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy-\int\frac{1 -\frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy)\] Consider \[\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\] Let u = \(y-\frac{1}{y}\) \(du =(1 + \frac{1}{y^2}) dy\) and \(y^2 + 1/y^2 = (y-\frac{1}{y})^2+2=u^2+2\) So, \[\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\]\[=\int \frac{1}{u^2+2}du\] Similarly, for\(\int \frac{1-\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\) Let \(v = y+ \frac{1}{y}\) \[dv = 1 - \frac{1}{y^2}dy\]\[y^2 + 1/y^2 = (y+\frac{1}{y})^2-2=v^2-2\] So, \[\int \frac{1-\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\]\[=\int\frac{1}{v^2-2}dv\] As a result, \[\frac{1}{2}(\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy-\int\frac{1 -\frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy)\]\[=\frac{1}{2}(\int\frac{1}{u^2+2}du-\int\frac{1}{v^2-2}dv)\]And You can integrate it then.
primeralph
  • primeralph
@roly Thank you for taking the time to illustrate it.

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