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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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

can you tell me how to at least start it?

- 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

does anyone know of a completed example of a problem like this I can go online and review or anything?

- primeralph

|dw:1368859626389:dw|

- primeralph

That's how you start.

- anonymous

can you tell me what the J is for? I have never ever seen that used before

- primeralph

|dw:1368859717310:dw|

- primeralph

J is just a multiplier. In this case, J = 1 , so you can remove it

- anonymous

(x^8+1) means: X^9

- anonymous

I highly doubt that it would be that simple felavin.

- anonymous

WHY? SMOKEYDABEAR.

- anonymous

could the next step be to let u=tan(theta) then du=sec^2(theta) d(theta)

- primeralph

@SinginDaCalc2Blues same thing.........you're looping

- anonymous

Yeah all that would do is put you right back to where you started

- primeralph

^^^ that girl is just too smart..........

- anonymous

Too stupid*

- anonymous

I'm just here to learn how to do it myself

- anonymous

are you people calling me stupid? seriously?

- anonymous

No, I was calling myself stupid xD

- primeralph

no, I didn't mean that.

- primeralph

okay, let's figure out an easier way to solve this problem.

- primeralph

@SinginDaCalc2Blues can you handle recursive integrals?

- 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

@SinginDaCalc2Blues we really would never do that.

- 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

wait............a minute...why not just polar?

- anonymous

Polar coordinates? Actually that makes some sense...

- anonymous

are you saying turn the integral into a polar expression or something?

- primeralph

@smokeydabear tricky thing is, it returns the integrand to the form we had before....because using cos, cos = 1/sec.....

- anonymous

my professor is devious!

- anonymous

Your professor is pretty evil for giving you something like this

- anonymous

would u=x^2
du=2x dx
indefinite integral of 1/(u^4+1)
help?
(sorry, my latex isn't working)

- primeralph

@SinginDaCalc2Blues I reduced it to......can you solve this?|dw:1368860855998:dw|

- anonymous

the int, called complex in

- anonymous

uhm if k is a constant, what is J again?

- primeralph

J is the variable of integration, just like U

- primeralph

U in U substitution

- anonymous

I just looked it up on wolfram... what is this monstrosity...

- primeralph

@smokeydabear I told you...........

- anonymous

How do you know where to even start with something like this? That's always been my problem

- anonymous

a good place to start is with Vodka :o)

- primeralph

@SinginDaCalc2Blues I asked you if you knew recursives because as you can see here, the trigs are repeating..which means recursion

##### 1 Attachment

- anonymous

no, not vodka, start with despair.

- anonymous

in 30 minutes i will no longer be able to do integration but I will be able to slurrrr Russian! :o)~

- anonymous

vodka inhibits your ability to do math.

- anonymous

despair only temporarily inhibits your ability to do everything

- primeralph

@smokeydabear recursion works to start. for example|dw:1368861204284:dw|

- anonymous

Okay, so how would we use that to help us? I'm still confused :s

- anonymous

crazy problem o:

- anonymous

well long only :P

- primeralph

@smokeydabear it saves time because once you know that it will repeat, you can generalize..........

- anonymous

hey guys...i have an idea...gimme a sec...

- primeralph

.........meaning............|dw:1368861457508:dw|

- stamp

u = x^2, du = 2x dx\[\int \frac{du}{u^4+1}\]

- primeralph

so, because they occur in trig functions, we can use it for the tan theta

- primeralph

@stamp did that

- anonymous

Ohhhhhh I see what you did with the sin's and the pi's @primeralph very clever! =D

- anonymous

now we just need to learn to use it effectively...

- primeralph

|dw:1368861979382:dw|

- anonymous

I gotcha Ralph, I noticed it too but I didn't want to be mean =P

- anonymous

##### 1 Attachment

- anonymous

Does this little bit of trickyness open any doors?

- anonymous

yeah what stamp wrote I believe is wrong...it wouldn't be x^4...it would be (x^4)^2

- anonymous

Well what stamp wrote isn't technically wrong, but it isn't the method to solving this.

- anonymous

I basically factored out of the denominator an x^2...then did u-sub

- anonymous

##### 1 Attachment

- anonymous

hi, the integrate is simple: Who can solve this integrate:\[\int\limits_{}^{} \frac{\tan Z }{ Z- \frac{ P }{ 4}}\] GAME IS OVER ..........

- anonymous

hello

- 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

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

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

##### 1 Attachment

- anonymous

Then \[u^3+1=(u+1)(u^2-u+1)\], use partial fractions.

- anonymous

what do i do with the numerator?

- 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

so then...after I do that, will i have technically several different integrals which will be easier to integrate?

- 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

i havent read much what is written above, its too long, has the question been solved ?

- anonymous

ok drawar...the answer definitely has some nasty ln's...i will keep plugging away at it...one more question...

- 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

has anyone tried x^2 =t ?

- anonymous

yeah shub...wayyy up at the top...we are past that now...but thanks

- anonymous

Yes @SinginDaCalc2Blues

- 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

ok...i will keep plugging away and see if I can get it...this problem has been so crazy! :o)

- anonymous

are you using imaginary numbers shub?

- anonymous

is it a definite/indefinite integral in your original question?

- shubhamsrg

yes, the very same.
i can be treated as just another constant

- anonymous

indefinite

- anonymous

i have no idea how to use imaginary numbers...i'm only in calc 2..we haven't used those

- 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

I wish I had learned about them...i just hope i can solve this

- 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

@roly Thank you for taking the time to illustrate it.

Looking for something else?

Not the answer you are looking for? Search for more explanations.