anonymous
  • anonymous
How to integrate this
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
|dw:1356849576104:dw|
hartnn
  • hartnn
seems it doesn't have any closed form... http://www.wolframalpha.com/input/?i=integral+%28x%5E%281%2Fn%29-1%2Fx%5En%29%5En
abb0t
  • abb0t
are we assuming n is a constant?

Looking for something else?

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

More answers

anonymous
  • anonymous
\[=\int\limits \left( \frac{x^{\frac{n^2+1}{n}}-1}{x^n} \right)^ndx\]
anonymous
  • anonymous
I 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=(...)^(n-1), 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.
anonymous
  • anonymous
@abb0t We know \(n\) is constant... but it would help to know if \(n\) is just a natural number or any real number.
anonymous
  • anonymous
For some reason things like this want me to try binomial theorem, lol I'm just a bit crazy I guess.
abb0t
  • abb0t
crazy
anonymous
  • anonymous
If you assume \(n \in \mathbb{N}\) you can just use binomial theorem and get an answer in summation notation.
anonymous
  • anonymous
Then if you're lucky, you might be able to simplify it from there, into something algebraic, but there is no guarantee it will happen.
anonymous
  • anonymous
There is no reason to be afraid of summations when it comes to certain integrals... there is no guarantee the function isn't transcendental.
anonymous
  • anonymous
=\[\int\limits \sum_{k=0}^{\infty} \left(\begin{matrix}n \\ k\end{matrix}\right)(x^{\frac{1}{n}})^{n-k}x^{-nk}dx\] 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.
anonymous
  • anonymous
I would assume it doesn't matter.
anonymous
  • anonymous
And that last x should have an (-1)^k also.

Looking for something else?

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