A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Hi! Integrate 2x/(x^8+1) dx
anonymous
 3 years ago
Hi! Integrate 2x/(x^8+1) dx

This Question is Closed

anonymous
 3 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

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

anonymous
 3 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.

anonymous
 3 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?

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

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

anonymous
 3 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

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

anonymous
 3 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

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

anonymous
 3 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)

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

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

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

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

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

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

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

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

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

anonymous
 3 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...

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

anonymous
 3 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?

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

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

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

anonymous
 3 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.....

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

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

anonymous
 3 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)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the int, called complex in

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

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

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

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

anonymous
 3 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

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

anonymous
 3 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

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

anonymous
 3 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)~

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

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

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

anonymous
 3 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

anonymous
 3 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..........

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

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

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

anonymous
 3 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

anonymous
 3 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

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

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

anonymous
 3 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

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

anonymous
 3 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

anonymous
 3 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.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hi, the integrate is simple: Who can solve this integrate:\[\int\limits_{}^{} \frac{\tan Z }{ Z \frac{ P }{ 4}}\] GAME IS OVER ..........

anonymous
 3 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)

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

anonymous
 3 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?

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

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

anonymous
 3 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!

anonymous
 3 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?

anonymous
 3 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
 3 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 ?

anonymous
 3 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...

anonymous
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0has anyone tried x^2 =t ?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes @SinginDaCalc2Blues

shubhamsrg
 3 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

anonymous
 3 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)

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

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

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

anonymous
 3 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
 3 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 (:

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\frac{2x}{x^8+1}dx\]\[=\int \frac{1}{x^8+1}d(x^2)\]\[=\int \frac{1}{y^4+1}dy\]\[=\frac{1}{2}\int \frac{y^2+1y^2 +1}{y^4+1} dy\]\[=\frac{1}{2}(\int \frac{y^2+1}{y^4+1}dy\int\frac{y^2 1}{y^4+1} dy)\]\[=\frac{1}{2}(\int \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\int\frac{1 \frac{1}{y^2}}{y^2+\frac{1}{y^2}} dy)\] 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@roly Thank you for taking the time to illustrate it.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.