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primeralphBest ResponseYou've already chosen the best response.0
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
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
can you tell me how to at least start it?
 11 months ago

smokeydabearBest ResponseYou've already chosen the best response.0
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.
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
does anyone know of a completed example of a problem like this I can go online and review or anything?
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
dw:1368859626389:dw
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
That's how you start.
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
can you tell me what the J is for? I have never ever seen that used before
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
dw:1368859717310:dw
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
J is just a multiplier. In this case, J = 1 , so you can remove it
 11 months ago

smokeydabearBest ResponseYou've already chosen the best response.0
I highly doubt that it would be that simple felavin.
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
could the next step be to let u=tan(theta) then du=sec^2(theta) d(theta)
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
@SinginDaCalc2Blues same thing.........you're looping
 11 months ago

smokeydabearBest ResponseYou've already chosen the best response.0
Yeah all that would do is put you right back to where you started
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
^^^ that girl is just too smart..........
 11 months ago

smokeydabearBest ResponseYou've already chosen the best response.0
I'm just here to learn how to do it myself
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
are you people calling me stupid? seriously?
 11 months ago

smokeydabearBest ResponseYou've already chosen the best response.0
No, I was calling myself stupid xD
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
no, I didn't mean that.
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
okay, let's figure out an easier way to solve this problem.
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
@SinginDaCalc2Blues can you handle recursive integrals?
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
okay...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...
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
@SinginDaCalc2Blues we really would never do that.
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
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?
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
wait............a minute...why not just polar?
 11 months ago

smokeydabearBest ResponseYou've already chosen the best response.0
Polar coordinates? Actually that makes some sense...
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
are you saying turn the integral into a polar expression or something?
 11 months ago

primeralphBest 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.....
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
my professor is devious!
 11 months ago

smokeydabearBest ResponseYou've already chosen the best response.0
Your professor is pretty evil for giving you something like this
 11 months ago

Peter14Best ResponseYou've already chosen the best response.0
would u=x^2 du=2x dx indefinite integral of 1/(u^4+1) help? (sorry, my latex isn't working)
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
@SinginDaCalc2Blues I reduced it to......can you solve this?dw:1368860855998:dw
 11 months ago

felavinBest ResponseYou've already chosen the best response.1
the int, called complex in
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
uhm if k is a constant, what is J again?
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
J is the variable of integration, just like U
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
U in U substitution
 11 months ago

smokeydabearBest ResponseYou've already chosen the best response.0
I just looked it up on wolfram... what is this monstrosity...
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
@smokeydabear I told you...........
 11 months ago

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

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
a good place to start is with Vodka :o)
 11 months ago

primeralphBest 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
 11 months ago

Peter14Best ResponseYou've already chosen the best response.0
no, not vodka, start with despair.
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
in 30 minutes i will no longer be able to do integration but I will be able to slurrrr Russian! :o)~
 11 months ago

Peter14Best ResponseYou've already chosen the best response.0
vodka inhibits your ability to do math.
 11 months ago

Peter14Best ResponseYou've already chosen the best response.0
despair only temporarily inhibits your ability to do everything
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
@smokeydabear recursion works to start. for exampledw:1368861204284:dw
 11 months ago

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

some_someoneBest ResponseYou've already chosen the best response.2
crazy problem o:
 11 months ago

some_someoneBest ResponseYou've already chosen the best response.2
well long only :P
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
@smokeydabear it saves time because once you know that it will repeat, you can generalize..........
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
hey guys...i have an idea...gimme a sec...
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
.........meaning............dw:1368861457508:dw
 11 months ago

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

primeralphBest ResponseYou've already chosen the best response.0
so, because they occur in trig functions, we can use it for the tan theta
 11 months ago

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

Peter14Best ResponseYou've already chosen the best response.0
now we just need to learn to use it effectively...
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
dw:1368861979382:dw
 11 months ago

smokeydabearBest ResponseYou've already chosen the best response.0
I gotcha Ralph, I noticed it too but I didn't want to be mean =P
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
Does this little bit of trickyness open any doors?
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
yeah what stamp wrote I believe is wrong...it wouldn't be x^4...it would be (x^4)^2
 11 months ago

smokeydabearBest ResponseYou've already chosen the best response.0
Well what stamp wrote isn't technically wrong, but it isn't the method to solving this.
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
I basically factored out of the denominator an x^2...then did usub
 11 months ago

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

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
hi...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)
 11 months ago

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

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
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?
 11 months ago

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

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
what do i do with the numerator?
 11 months ago

drawarBest 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!
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
so then...after I do that, will i have technically several different integrals which will be easier to integrate?
 11 months ago

drawarBest ResponseYou've already chosen the best response.0
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.
 11 months ago

shubhamsrgBest ResponseYou've already chosen the best response.0
i havent read much what is written above, its too long, has the question been solved ?
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
ok drawar...the answer definitely has some nasty ln's...i will keep plugging away at it...one more question...
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
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?
 11 months ago

shubhamsrgBest ResponseYou've already chosen the best response.0
has anyone tried x^2 =t ?
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
yeah shub...wayyy up at the top...we are past that now...but thanks
 11 months ago

drawarBest ResponseYou've already chosen the best response.0
Yes @SinginDaCalc2Blues
 11 months ago

shubhamsrgBest ResponseYou've already chosen the best response.0
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
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
ok...i will keep plugging away and see if I can get it...this problem has been so crazy! :o)
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
are you using imaginary numbers shub?
 11 months ago

drawarBest ResponseYou've already chosen the best response.0
is it a definite/indefinite integral in your original question?
 11 months ago

shubhamsrgBest ResponseYou've already chosen the best response.0
yes, the very same. i can be treated as just another constant
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
indefinite
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
i have no idea how to use imaginary numbers...i'm only in calc 2..we haven't used those
 11 months ago

shubhamsrgBest ResponseYou've already chosen the best response.0
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 (:
 11 months ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
I wish I had learned about them...i just hope i can solve this
 11 months ago

RolyPolyBest 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.
 11 months ago

primeralphBest ResponseYou've already chosen the best response.0
@roly Thank you for taking the time to illustrate it.
 11 months ago
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