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SinginDaCalc2Blues
 2 years ago
Hi! Integrate 2x/(x^8+1) dx
SinginDaCalc2Blues
 2 years ago
Hi! Integrate 2x/(x^8+1) dx

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primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0I'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

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0can you tell me how to at least start it?

smokeydabear
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0does anyone know of a completed example of a problem like this I can go online and review or anything?

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1368859626389:dw

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0That's how you start.

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0can you tell me what the J is for? I have never ever seen that used before

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1368859717310:dw

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0J is just a multiplier. In this case, J = 1 , so you can remove it

smokeydabear
 2 years ago
Best ResponseYou've already chosen the best response.0I highly doubt that it would be that simple felavin.

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0could the next step be to let u=tan(theta) then du=sec^2(theta) d(theta)

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0@SinginDaCalc2Blues same thing.........you're looping

smokeydabear
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah all that would do is put you right back to where you started

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0^^^ that girl is just too smart..........

smokeydabear
 2 years ago
Best ResponseYou've already chosen the best response.0I'm just here to learn how to do it myself

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0are you people calling me stupid? seriously?

smokeydabear
 2 years ago
Best ResponseYou've already chosen the best response.0No, I was calling myself stupid xD

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0no, I didn't mean that.

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0okay, let's figure out an easier way to solve this problem.

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0@SinginDaCalc2Blues can you handle recursive integrals?

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0okay...well I appreciate the help from everyone but it's hard to understand the context of nonverbal communication sometimes...I was hoping you guys weren't calling me stupid...anyway...

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0@SinginDaCalc2Blues we really would never do that.

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0i 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
 2 years ago
Best ResponseYou've already chosen the best response.0wait............a minute...why not just polar?

smokeydabear
 2 years ago
Best ResponseYou've already chosen the best response.0Polar coordinates? Actually that makes some sense...

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0are you saying turn the integral into a polar expression or something?

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0@smokeydabear tricky thing is, it returns the integrand to the form we had before....because using cos, cos = 1/sec.....

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0my professor is devious!

smokeydabear
 2 years ago
Best ResponseYou've already chosen the best response.0Your professor is pretty evil for giving you something like this

Peter14
 2 years ago
Best ResponseYou've already chosen the best response.0would u=x^2 du=2x dx indefinite integral of 1/(u^4+1) help? (sorry, my latex isn't working)

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0@SinginDaCalc2Blues I reduced it to......can you solve this?dw:1368860855998:dw

felavin
 2 years ago
Best ResponseYou've already chosen the best response.1the int, called complex in

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0uhm if k is a constant, what is J again?

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0J is the variable of integration, just like U

smokeydabear
 2 years ago
Best ResponseYou've already chosen the best response.0I just looked it up on wolfram... what is this monstrosity...

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0@smokeydabear I told you...........

smokeydabear
 2 years ago
Best ResponseYou've already chosen the best response.0How do you know where to even start with something like this? That's always been my problem

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0a good place to start is with Vodka :o)

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0@SinginDaCalc2Blues I asked you if you knew recursives because as you can see here, the trigs are repeating..which means recursion

Peter14
 2 years ago
Best ResponseYou've already chosen the best response.0no, not vodka, start with despair.

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0in 30 minutes i will no longer be able to do integration but I will be able to slurrrr Russian! :o)~

Peter14
 2 years ago
Best ResponseYou've already chosen the best response.0vodka inhibits your ability to do math.

Peter14
 2 years ago
Best ResponseYou've already chosen the best response.0despair only temporarily inhibits your ability to do everything

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0@smokeydabear recursion works to start. for exampledw:1368861204284:dw

smokeydabear
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, so how would we use that to help us? I'm still confused :s

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0@smokeydabear it saves time because once you know that it will repeat, you can generalize..........

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0hey guys...i have an idea...gimme a sec...

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0.........meaning............dw:1368861457508:dw

stamp
 2 years ago
Best ResponseYou've already chosen the best response.0u = x^2, du = 2x dx\[\int \frac{du}{u^4+1}\]

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0so, because they occur in trig functions, we can use it for the tan theta

smokeydabear
 2 years ago
Best ResponseYou've already chosen the best response.0Ohhhhhh I see what you did with the sin's and the pi's @primeralph very clever! =D

Peter14
 2 years ago
Best ResponseYou've already chosen the best response.0now we just need to learn to use it effectively...

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1368861979382:dw

smokeydabear
 2 years ago
Best ResponseYou've already chosen the best response.0I gotcha Ralph, I noticed it too but I didn't want to be mean =P

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0Does this little bit of trickyness open any doors?

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0yeah what stamp wrote I believe is wrong...it wouldn't be x^4...it would be (x^4)^2

smokeydabear
 2 years ago
Best ResponseYou've already chosen the best response.0Well what stamp wrote isn't technically wrong, but it isn't the method to solving this.

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0I basically factored out of the denominator an x^2...then did usub

felavin
 2 years ago
Best ResponseYou've already chosen the best response.1hi, the integrate is simple: Who can solve this integrate:\[\int\limits_{}^{} \frac{\tan Z }{ Z \frac{ P }{ 4}}\] GAME IS OVER ..........

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0hi...i'm only in calc 2...i don't know what CauchyReimann equations are... I did however get my integral down to 1/(u^3+1)

drawar
 2 years ago
Best ResponseYou've already chosen the best response.0In case you love twisting with partial fractions: \[\frac{ 2x }{ x^8 +1}=\frac{ 2x }{( x^4\sqrt{2}x^2+1)( x^4+\sqrt{2}x^2+1)}\]

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0hi 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?

drawar
 2 years ago
Best ResponseYou've already chosen the best response.0Then \[u^3+1=(u+1)(u^2u+1)\], use partial fractions.

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0what do i do with the numerator?

drawar
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ u^3+1 }=\frac{ A }{ u+1 }+\frac{ Bu+C }{ u^2u+1 }\] =>\[A(u^2u+1)+(Bu+C)(u+1)=1\] Then equate the coefficients and you'll get a system of linear equations in terms of A,B, and C. Solve it!

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0so then...after I do that, will i have technically several different integrals which will be easier to integrate?

drawar
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 2 years ago
Best ResponseYou've already chosen the best response.0i havent read much what is written above, its too long, has the question been solved ?

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0ok drawar...the answer definitely has some nasty ln's...i will keep plugging away at it...one more question...

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0once 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
 2 years ago
Best ResponseYou've already chosen the best response.0has anyone tried x^2 =t ?

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0yeah shub...wayyy up at the top...we are past that now...but thanks

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0well, 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

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0ok...i will keep plugging away and see if I can get it...this problem has been so crazy! :o)

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0are you using imaginary numbers shub?

drawar
 2 years ago
Best ResponseYou've already chosen the best response.0is it a definite/indefinite integral in your original question?

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0yes, the very same. i can be treated as just another constant

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0i have no idea how to use imaginary numbers...i'm only in calc 2..we haven't used those

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0I 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 (:

SinginDaCalc2Blues
 2 years ago
Best ResponseYou've already chosen the best response.0I wish I had learned about them...i just hope i can solve this

RolyPoly
 2 years ago
Best ResponseYou've already chosen the best response.1\[\int\frac{2x}{x^8+1}dx\]\[=\int \frac{1}{x^8+1}d(x^2)\]\[=\int \frac{1}{y^4+1}dy\]\[=\frac{1}{2}\int \frac{y^2+1y^2 +1}{y^4+1} dy\]\[=\frac{1}{2}(\int \frac{y^2+1}{y^4+1}dy\int\frac{y^2 1}{y^4+1} dy)\]\[=\frac{1}{2}(\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\int\frac{1 \frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy)\] 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 \(v = y+ \frac{1}{y}\) \[dv = 1  \frac{1}{y^2}dy\]\[y^2 + 1/y^2 = (y+\frac{1}{y})^22=v^22\] So, \[\int \frac{1\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\]\[=\int\frac{1}{v^22}dv\] As a result, \[\frac{1}{2}(\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\int\frac{1 \frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy)\]\[=\frac{1}{2}(\int\frac{1}{u^2+2}du\int\frac{1}{v^22}dv)\]And You can integrate it then.

primeralph
 2 years ago
Best ResponseYou've already chosen the best response.0@roly Thank you for taking the time to illustrate it.
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