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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

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    $\begingroup$ Are you familiar with the method of partial fractions? Or, if you're given that the indefinite integral is of that form, you can just take the derivative of the right-hand side. $\endgroup$ – Robert Shore Mar 16 at 2:17
  • $\begingroup$ @Robert taking partial fraction is very lengthy I am trying to solve it less lengthy way $\endgroup$ – jacky Mar 16 at 2:21
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    $\begingroup$ I don't see any short cuts beyond noticing that the denominator of the fraction is $(x^5-1)^2$. $\endgroup$ – Robert Shore Mar 16 at 2:26
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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 $$px^3+qx^8=-x^3(x^5-1)=x^3-x^8\implies p=1 \qquad \text{and} \qquad q=-1$$

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\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}

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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: $$\begin{cases}q=-1\\ 7p+6q=1 \\ p+2q=-1\\ 3p=3\end{cases}\implies \bbox [5px,border:2px solid #C0A000]{\begin{array}pp=+1 \\ q=-1\end{array}}$$

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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 $\displaystyle \int t^{-2}dt=-\frac{1}{t}+C=-\frac{x^3}{x^5-1}+C$

$$\int\frac{2x^7+3x^2}{(x^5-1)^2}dx=\frac{x^3-x^8}{(x^5-1)^2}+C$$

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