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 
 A: 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
$$px^3+qx^8=-x^3(x^5-1)=x^3-x^8\implies p=1 \qquad \text{and} \qquad q=-1$$
A: \begin{equation}
\left(\frac{px^3+qx^8}{x^{10}-2x^5+1}\right)^\prime=\frac{(8qx^7+3px^2)(x^5-1)^2-10x^7(qx^5+p)(x^5-1)}{(x^5-1)^4}
\end{equation}
Let $p=1,\quad q=-1$ and we get
\begin{eqnarray}
&&\frac{(-8x^7+3x^2)(x^5-1)^2-10x^7(-x^5+1)(x^5-1)}{(x^5-1)^4}\\
&=&\frac{(-8x^7+3x^2)(x^5-1)^2+10x^7(x^5-1)^2}{(x^5-1)^4}\\
&=&\frac{2x^7+3x^2}{(x^5-1)^2}
\end{eqnarray}
A: The best way I can seem to think of is differentiating both sides of the equality that gives you the following: $$\dfrac{\mathrm d}{\mathrm dx}\int\dfrac{2x^7+3x^2}{x^{10}-2x^5+1}\mathrm dx=\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac{px^3+qx^8}{x^{10}-2x^5+1}\right)$$ 

$$ \begin{equation}\begin{aligned} \dfrac{\mathrm d}{\mathrm dx}\left(\dfrac{px^3+qx^8}{x^{10}-2x^5+1}\right) &= \dfrac{(3px^{2}+8qx^7)(x^{10}-2x^5+1)-(10x^9-10x^4)(px^3+qx^8)}{(x^{10}-2x^5+1)^2} \\ &= \dfrac{-2qx^{17}-7px^{12}-6qx^{12}+4px^7+8qx^7+3px^2}{(x^{10}-2x^5+1)^2}\end{aligned}\tag1\end{equation}$$

$$\begin{equation}\begin{aligned}\dfrac{\mathrm d}{\mathrm dx}\int\dfrac{2x^7+3x^2}{x^{10}-2x^5+1}\mathrm dx&=\dfrac{2x^7+3x^2}{x^{10}-2x^5+1}\\&=\dfrac{(2x^7+3x^2)(x^{10}-2x^5+1)}{(x^{10}-2x^5+1)^2}\\&=\dfrac{2x^{17}-x^{12}-4x^7+3x^2}{(x^{10}-2x^5+1)^2}\end{aligned}\end{equation}\tag2$$

Equating $(1)$ and $(2)$ and comparing gives us the following system: $$\begin{cases}q=-1\\ 7p+6q=1 \\ p+2q=-1\\ 3p=3\end{cases}\implies \bbox [5px,border:2px solid #C0A000]{\begin{array}pp=+1 \\ q=-1\end{array}}$$
A: thanks friends got the result
$\displaystyle \int\frac{2x^7+3x^2}{(x^5-1)^2}dx=\int\frac{2x+3x^{-4}}{(x^2-x^{-3})^2}dx$
put $x^2-x^{-3}=t$ and $(2x+3x^{-4})dx=dt$
Integration is $\displaystyle \int t^{-2}dt=-\frac{1}{t}+C=-\frac{x^3}{x^5-1}+C$
$$\int\frac{2x^7+3x^2}{(x^5-1)^2}dx=\frac{x^3-x^8}{(x^5-1)^2}+C$$
