## sauravshakya 2 years ago How to integrate this

1. sauravshakya

|dw:1356849576104:dw|

2. 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

3. abb0t

are we assuming n is a constant?

4. malevolence19

$=\int\limits \left( \frac{x^{\frac{n^2+1}{n}}-1}{x^n} \right)^ndx$

5. malevolence19

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.

6. wio

@abb0t We know $$n$$ is constant... but it would help to know if $$n$$ is just a natural number or any real number.

7. wio

For some reason things like this want me to try binomial theorem, lol I'm just a bit crazy I guess.

8. abb0t

crazy

9. wio

If you assume $$n \in \mathbb{N}$$ you can just use binomial theorem and get an answer in summation notation.

10. wio

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.

11. wio

There is no reason to be afraid of summations when it comes to certain integrals... there is no guarantee the function isn't transcendental.

12. malevolence19

=$\int\limits \sum_{k=0}^{\infty} \left(\begin{matrix}n \\ k\end{matrix}\right)(x^{\frac{1}{n}})^{n-k}x^{-nk}dx$ 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.

13. malevolence19

I would assume it doesn't matter.

14. malevolence19

And that last x should have an (-1)^k also.