Integral as a Convergent Series of Real Numbers

[Seeker555]

Find
\(\displaystyle \int \frac{1}{1+x^6}dx\) as a convergent series of real numbers.

The boundaries for the integral are 1/3 and -1/3 for the upper and lower (obviously hehe). Sorry i couldn't find the code input for boundaries on the integral.

I'm pretty clueless so any help is great.

 

[Seeker555]

any help?

 

[pka]

Find
\(\displaystyle \int \frac{1}{1+x^6}dx\) as a convergent series of real numbers.

any help?

Note on the LaTeX. [TEX]\int\limits_{-1/3}^{1/3} \dfrac{dx}{1+x^6}[/TEX] gives \(\displaystyle \int\limits_{-1/3}^{1/3} \dfrac{dx}{1+x^6}\)

Note also you have an even function so \(\displaystyle \int\limits_{-1/3}^{1/3} \dfrac{dx}{1+x^6}=2\int\limits_{0}^{1/3} \dfrac{dx}{1+x^6}\)
That simplifies the question.

I wish I could help you. But I don't really understand what the person who set this question expects. If you could post any lecture notes on this topic maybe someone here could see what is expected.

 

[daon2]

It looks like you are meant to find a power series for your integrand, and then integrate?

\(\displaystyle \frac{1}{1+x^6} = \frac{1}{1-(-x^6)}\)

Now recall

\(\displaystyle \frac{1}{1-t} = \displaystyle \sum_{n=0}^{\infty}t^n\)

Integrate the power series now, term by term. It will converge for +/- 1/3.