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Hi! Integrate 2x/(x^8+1) dx

Calculus1
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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
can you tell me how to at least start it?
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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does anyone know of a completed example of a problem like this I can go online and review or anything?
|dw:1368859626389:dw|
That's how you start.
can you tell me what the J is for? I have never ever seen that used before
|dw:1368859717310:dw|
J is just a multiplier. In this case, J = 1 , so you can remove it
(x^8+1) means: X^9
I highly doubt that it would be that simple felavin.
WHY? SMOKEYDABEAR.
could the next step be to let u=tan(theta) then du=sec^2(theta) d(theta)
@SinginDaCalc2Blues same thing.........you're looping
Yeah all that would do is put you right back to where you started
^^^ that girl is just too smart..........
Too stupid*
I'm just here to learn how to do it myself
are you people calling me stupid? seriously?
No, I was calling myself stupid xD
no, I didn't mean that.
okay, let's figure out an easier way to solve this problem.
@SinginDaCalc2Blues can you handle recursive integrals?
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...
@SinginDaCalc2Blues we really would never do that.
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?
wait............a minute...why not just polar?
Polar coordinates? Actually that makes some sense...
are you saying turn the integral into a polar expression or something?
@smokeydabear tricky thing is, it returns the integrand to the form we had before....because using cos, cos = 1/sec.....
my professor is devious!
Your professor is pretty evil for giving you something like this
would u=x^2 du=2x dx indefinite integral of 1/(u^4+1) help? (sorry, my latex isn't working)
@SinginDaCalc2Blues I reduced it to......can you solve this?|dw:1368860855998:dw|
the int, called complex in
uhm if k is a constant, what is J again?
J is the variable of integration, just like U
U in U substitution
I just looked it up on wolfram... what is this monstrosity...
@smokeydabear I told you...........
How do you know where to even start with something like this? That's always been my problem
a good place to start is with Vodka :o)
@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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no, not vodka, start with despair.
in 30 minutes i will no longer be able to do integration but I will be able to slurrrr Russian! :o)~
vodka inhibits your ability to do math.
despair only temporarily inhibits your ability to do everything
@smokeydabear recursion works to start. for example|dw:1368861204284:dw|
Okay, so how would we use that to help us? I'm still confused :s
crazy problem o:
well long only :P
@smokeydabear it saves time because once you know that it will repeat, you can generalize..........
hey guys...i have an idea...gimme a sec...
.........meaning............|dw:1368861457508:dw|
u = x^2, du = 2x dx\[\int \frac{du}{u^4+1}\]
so, because they occur in trig functions, we can use it for the tan theta
@stamp did that
Ohhhhhh I see what you did with the sin's and the pi's @primeralph very clever! =D
now we just need to learn to use it effectively...
|dw:1368861979382:dw|
I gotcha Ralph, I noticed it too but I didn't want to be mean =P
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Does this little bit of trickyness open any doors?
yeah what stamp wrote I believe is wrong...it wouldn't be x^4...it would be (x^4)^2
Well what stamp wrote isn't technically wrong, but it isn't the method to solving this.
I basically factored out of the denominator an x^2...then did u-sub
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hi, the integrate is simple: Who can solve this integrate:\[\int\limits_{}^{} \frac{\tan Z }{ Z- \frac{ P }{ 4}}\] GAME IS OVER ..........
hello
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)
In case you love twisting with partial fractions: \[\frac{ 2x }{ x^8 +1}=\frac{ 2x }{( x^4-\sqrt{2}x^2+1)( x^4+\sqrt{2}x^2+1)}\]
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?
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Then \[u^3+1=(u+1)(u^2-u+1)\], use partial fractions.
what do i do with the numerator?
\[\frac{ 1 }{ u^3+1 }=\frac{ A }{ u+1 }+\frac{ Bu+C }{ u^2-u+1 }\] =>\[A(u^2-u+1)+(Bu+C)(u+1)=1\] Then equate the coefficients and you'll get a system of linear equations in terms of A,B, and C. Solve it!
so then...after I do that, will i have technically several different integrals which will be easier to integrate?
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.
i havent read much what is written above, its too long, has the question been solved ?
ok drawar...the answer definitely has some nasty ln's...i will keep plugging away at it...one more question...
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?
has anyone tried x^2 =t ?
yeah shub...wayyy up at the top...we are past that now...but thanks
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
ok...i will keep plugging away and see if I can get it...this problem has been so crazy! :o)
are you using imaginary numbers shub?
is it a definite/indefinite integral in your original question?
yes, the very same. i can be treated as just another constant
indefinite
i have no idea how to use imaginary numbers...i'm only in calc 2..we haven't used those
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 (:
I wish I had learned about them...i just hope i can solve this
\[\int\frac{2x}{x^8+1}dx\]\[=\int \frac{1}{x^8+1}d(x^2)\]\[=\int \frac{1}{y^4+1}dy\]\[=\frac{1}{2}\int \frac{y^2+1-y^2 +1}{y^4+1} dy\]\[=\frac{1}{2}(\int \frac{y^2+1}{y^4+1}dy-\int\frac{y^2 -1}{y^4+1} dy)\]\[=\frac{1}{2}(\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy-\int\frac{1 -\frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy)\] 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.
@roly Thank you for taking the time to illustrate it.

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