Value of $(p,q)$ in indefinite integration Finding value of $(p,q)$ in  $\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx=\frac{px^3+qx^8}{x^{10}-2x^5+1}+c$
what i try
$\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx$
put $x=1/t$ and $dx=-1/t^2dt$
$\displaystyle -\frac{2t^5+3x^{10}}{t^{10}-2t^5+1}dt$
How do i solve it Help me please 

In fact, you are given that $$ \int\frac{2x^7+3x^2}{(x^5-1)^2}dx=\frac{px^3+qx^8}{(x^5-1)^2}+c$$ Because of the denominator in the integrand, rewrite the rhs as
$$\frac{P_n(x)}{x^5-1}$$ Differentiate both sides to get
$$\frac{2x^7+3x^2}{(x^5-1)^2}=\frac{\left(x^5-1\right) P_n'(x)-5 x^4 P_n(x)}{\left(x^5-1\right)^2}$$ that is to say
$${2x^7+3x^2}={\left(x^5-1\right) P_n'(x)-5 x^4 P_n(x)} $$In order to have $x^7$ implies that $5+(n-1)=7$ that is to say $n=3$. So, let
$$P_3(x)=a+bx+c x^2+dx^3$$ Replace, expand and group terms in the rhs to get
$${2x^7+3x^2}=-b-2 c x-3 d x^2-5 a x^4-4 b x^5-3 c x^6-2 d x^7$$ which implies $b=0$, $c=0$, $d=-1$ and $a=0$. So $P_3(x)=-x^3$. For here
$$px^3+qx^8=-x^3(x^5-1)=x^3-x^8\implies p=1 \qquad \text{and} \qquad q=-1$$
