Question on Indefinite Integration: $\int\frac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}\,\mathrm{d}x$ Give me some hints to start with this problem: $${\displaystyle\int}\dfrac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}\,\mathrm{d}x$$
 A: Because of the cube in denominator, you can say that $$\int \frac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}\,dx=\frac{P_n(x)}{(x^5+x^3+1)^2}$$ where $P_n(x)$ is a polynomial fo degree $n$.
Now differentiate both sides to get $$\frac{2x^{12}+5x^9}{\left(x^5+x^3+1\right)^3}=\frac{\left(x^5+x^3+1\right) P_n'(x)-2 x^2 \left(5 x^2+3\right)
   P_n(x)}{\left(x^5+x^3+1\right)^3}$$ So, the right hand side is of degree $n+4$; then, since the left hand side is of degree $12$, this implies $n=8$.
So, set $P_8(x)=\sum_{i=0}^8 a_i x^i$; replacing and comparing the terms of same power then gives the $a_i$'s.
Edit
This works but it is just junk when compared to the so beautiful solution given by André Nicolas.
A: HINT: $$x^5+x^3+1=x^5(1+x^{-2}+x^{-5})$$
Choose $1+x^{-2}+x^{-5}=u$
A: Let $x=1/t$. Then $dx=-dt/t^2$. Our integral becomes
$$ \int -\frac{2t+5t^4}{(1+t^2+t^5)^3}\,dt.$$
A: Hint 
Take $x^5$ common from denominator so that it's cube it will come out as $15$ and take $x^{15}$ common from  the numerator. Put the remaining denominator as $t$ numerator becomes $dt$. Hope you can do the rest.
