integrate dx/1+x^4

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integrate dx/1+x^4

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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already answered! r u an idiot?!
I think you need a review on the methods of trig substitution. You are asking for help with similar problems. Guys won't help you if they thing they are doing your homework for you.
sportsman u r mad ur mum retrice...illiterate boy

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this one is a little different because we can't u- sub it. I think it has to be done by partial fractions, and I'm not sure how to break it up exactly with that x^4 down there. I'm pretty sure it has to at least be Ax+B/(1+x^4), but I think there may be more with it.
It's a trig sub like all the rest have been. Let \(x^2 = tan\theta\)
Oneprince, the guys were a little harsh on you but don't get discouraged. This is the form of u substitution for tan inverse function.
The question is usually asked (x dx)/(1+x^4) in order to make the u sub work. Let u=x^2, du=2x, du/2=x. Becomes 1/2 integration du/1+u^2= 1/2 tan inverse u + C= 1/2 tan invers x^2 + C
ah hah, I knew it had to be something workable. Man, sometimes after a while these things slip you. In any event, do you mind putting the final answer you got, I'm just curious how you handled the dx portion of the u sub, and then getting it all back to x, letting x^2 = tan(theta). What a neat problem though. (for Polpak).
With integrals of inverse trig there is a cheat sheet that tells you integral of du/(a^2 + u^2) = (1/a) tan inverse (u+a) + C. An example integration of dx/(a^2+x^2) (a cannot be 0) Let u be x/a, du=dx/a Visualizing it, integration of [a(dx/a)]/a^2(1+(x/a)^2= 1/a integration (dx/a)/1+(x/a)^2 =(1/a) integration du/(1+u^2) =1/a tan inverse u + C =1/a tan inverse x/a + C Note: Oneprince problem has a higher exponent x^4. For this to work Oneprince problem must be xdx/1+x^4
that won't work on this one, because the dx, doesn't satisfy. You're right about the other one he had, it would work, but for this one, it does work out letting x^2 = tan(theta). I just wanted to see how he responded to it, because the algebra got a little tricky, and I'm always curious when things like this pop up. Normally I would have viewed it as an inverse tan deal, but the dx issue popped up, and then I thought maybe a partial fraction was needed, but couldn't resolve it, and when polpak made the ,it made sense. \[ \int\limits\limits_{?}^{?}(1/(1+\tan^2(\Theta))*(\sec^2(\theta)/2\sqrt{\tan(\theta)})d \Theta\] \[(1/2)\int\limits_{}^{}\sec^2(\theta)/(\tan(\theta)^(1/2)+\tan(\theta)^(5/2))d \theta\] let u = tan(theta) so then \[(1/2)\int\limits_{}^{}u^(-1/2)+u^(-5/2)du\], then it's all straight forward. I just wondered if polpak got down like that is all, and some techniques just don't alwasy pop back up when you've been doing something else for a while. I still think it's a dope eq.
Yeah I see that now.
I would bet it was a mistake on the part of the instructor. What is missing is an x upstairs. There are numerous questions like this in the inverse trig integral section.
he had the other version on here yesterday, then I came to this one for him, and left the partial bit, but polpak had the right idea. This is a perfect tig sub problem for the integration techniques section. That's why I love it, because it can throw you off a little, and it requires some manipulation too.

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