Let $f(x) = \int \frac{x}{1-x^{8}}dx\,$ 
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*Let $f(x) = \int \frac{x}{1-x^{8}}dx\,$.

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*Represent $I(x)$ by a power series $\sum^{\infty}a_{n}x^{n}$.(Find $a_{n}$)

*What is the radius of convergence of $I(x)$ ?


*Two curves are generated by polar equations $r=1+\sin\theta$ and $r=-\sin\theta$.

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*Find the area of the region that lies inside both two curves.

*Find the length of the part of the curve $r=1+\sin\theta$ that lies inside the curve $r=-\sin\theta$.



HINT:
For the first  problem, $$\frac x{1-x^8}=\frac x2\left(\frac1{1-x^4}+\frac1{1+x^4}\right)$$
$$=\frac x4\left(\frac1{1-x^2}+\frac1{1+x^2}\right)+\frac12\frac x{1+x^4}$$
$$=\frac x4\left(\frac1{1-x^2}\right)+\frac14\frac x{1+x^2}+\frac12\frac x{1+x^4}$$
$$=\frac 18\left(\frac{1+x-(1-x)}{1-x^2}\right)+\frac14\frac x{1+x^2}+\frac12\frac x{1+x^4}$$
$$=\frac 18\frac1{1-x}-\frac18\frac1{1+x}+\frac14\frac x{1+x^2}+\frac12\frac x{1+x^4}$$
For the last two terms put $x^2=u$
You will find all required the Series Formula here
In fact, we can make the substitution $x^2=u$ from the very start, but then we need substitute back $u$ with $x$ as we need the Power Series of $x$
