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sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1356849576104:dw

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1seems it doesn't have any closed form... http://www.wolframalpha.com/input/?i=integral+%28x%5E%281%2Fn%291%2Fx%5En%29%5En

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.0are we assuming n is a constant?

malevolence19
 2 years ago
Best ResponseYou've already chosen the best response.1\[=\int\limits \left( \frac{x^{\frac{n^2+1}{n}}1}{x^n} \right)^ndx\]

malevolence19
 2 years ago
Best ResponseYou've already chosen the best response.1I see 2 approaches. Either do something like: u=x^(n^2+1)/n  1 and solve for x in terms of u; then find dx, etc. Or try integration by parts by letting u=(...)^(n1), dv=(...)^1dx; from there you could try to establish a pattern for n. But it is more than likely unable to be solved in closed form.

wio
 2 years ago
Best ResponseYou've already chosen the best response.0@abb0t We know \(n\) is constant... but it would help to know if \(n\) is just a natural number or any real number.

wio
 2 years ago
Best ResponseYou've already chosen the best response.0For some reason things like this want me to try binomial theorem, lol I'm just a bit crazy I guess.

wio
 2 years ago
Best ResponseYou've already chosen the best response.0If you assume \(n \in \mathbb{N}\) you can just use binomial theorem and get an answer in summation notation.

wio
 2 years ago
Best ResponseYou've already chosen the best response.0Then if you're lucky, you might be able to simplify it from there, into something algebraic, but there is no guarantee it will happen.

wio
 2 years ago
Best ResponseYou've already chosen the best response.0There is no reason to be afraid of summations when it comes to certain integrals... there is no guarantee the function isn't transcendental.

malevolence19
 2 years ago
Best ResponseYou've already chosen the best response.1=\[\int\limits \sum_{k=0}^{\infty} \left(\begin{matrix}n \\ k\end{matrix}\right)(x^{\frac{1}{n}})^{nk}x^{nk}dx\] My only question with doing this is that in the binomial theorem the don't make mention of things of the form: \[(\alpha(n)+\beta(n))^n\] Where alpha and beta are numbers that depend on n.

malevolence19
 2 years ago
Best ResponseYou've already chosen the best response.1I would assume it doesn't matter.

malevolence19
 2 years ago
Best ResponseYou've already chosen the best response.1And that last x should have an (1)^k also.
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