Find the finite power series of $\frac{u^3+3u^2+3u+1}{u^2+2u+3}$

Question

I tried to do the partial fraction of $\frac{x^3}{(x-1)^4(x^2+2)}$. By letting $u=x-1$ I get $\frac{1}{u^4}\frac{u^3+3u^2+3u+1}{u^2+2u+3}$. I was able to get the answer by doing the long division of the $\frac{u^3+3u^2+3u+1}{u^2+2u+3}$ part. So, I wonder is it possible to turn $\frac{u^3+3u^2+3u+1}{u^2+2u+3}$ into a finite power series $a_{n}(x-c)^n $ with $c=0$ or turns $\frac{x^3}{(x^2+2)}$ into a finite power series $a_{n}(x-c)^n $ with $c=1$ so it can be divided by $(x-1)^4$ ?

Answer 1

$$\frac{u^3+3u^2+3u+1}{u^2+2u+3}=1+u-2\frac{u+1 }{u^2+2 u+3}$$ Now, let $$u^2+2u+3=(u-a)(u-b)\qquad\qquad a=-1-i \sqrt{2}\quad b=-1+i \sqrt{2}$$ $$\frac{u+1 }{u^2+2 u+3}=\frac{a+1}{(a-b) (u-a)}-\frac{b+1}{(a-b) (u-b)}$$ composing the series and playing with the complex numbers $$\frac{u^3+3u^2+3u+1}{u^2+2u+3}=\frac{1}{3}+\frac{7 }{9}u+\sum_{n=2}^\infty c_n\, u^n$$ with $$c_n=-\frac{\left(\sqrt{2}-2 i\right) \left(\frac{1}{3} i \left(\sqrt{2}+i\right)\right)^n+\left(\sqrt{2}+2 i\right) \left(-\frac{1}{3} i \left(\sqrt{2}-i\right)\right)^n}{3 \sqrt{2}}$$ which generate the sequence $$\left\{\frac{10}{27},-\frac{14}{81},-\frac{2}{243},\frac{46}{729},-\frac{86}{2187} ,\frac{34}{6561},\frac{190}{19683},-\frac{482}{59049},\frac{394}{177147}\right\}$$

Now, as your properly did, let $u=x-1$ to get $$\frac{x^3}{(x^2+2)}=$$ but this one is much simpler. Let $x^2=2t^2$ to make $$\sqrt{2} \frac{t^3}{1+t^2}$$ Use the series expansion of $\frac 1{1+y}$, replace $y$ by $t^2$ and multiply by $\sqrt{2} {t^3}$ and go backward from $t$ to $x$ and from $x$ to $u$.