# 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

• 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. – Robert Shore Mar 16 at 2:17
• @Robert taking partial fraction is very lengthy I am trying to solve it less lengthy way – jacky Mar 16 at 2:21
• I don't see any short cuts beyond noticing that the denominator of the fraction is $(x^5-1)^2$. – Robert Shore Mar 16 at 2:26

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 $$px^3+qx^8=-x^3(x^5-1)=x^3-x^8\implies p=1 \qquad \text{and} \qquad q=-1$$

$$$$\left(\frac{px^3+qx^8}{x^{10}-2x^5+1}\right)^\prime=\frac{(8qx^7+3px^2)(x^5-1)^2-10x^7(qx^5+p)(x^5-1)}{(x^5-1)^4}$$$$ Let $$p=1,\quad q=-1$$ and we get

$$\begin{eqnarray} &&\frac{(-8x^7+3x^2)(x^5-1)^2-10x^7(-x^5+1)(x^5-1)}{(x^5-1)^4}\\ &=&\frac{(-8x^7+3x^2)(x^5-1)^2+10x^7(x^5-1)^2}{(x^5-1)^4}\\ &=&\frac{2x^7+3x^2}{(x^5-1)^2} \end{eqnarray}$$

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{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

\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}\tag2

Equating $$(1)$$ and $$(2)$$ and comparing gives us the following system: $$\begin{cases}q=-1\\ 7p+6q=1 \\ p+2q=-1\\ 3p=3\end{cases}\implies \bbox [5px,border:2px solid #C0A000]{\begin{array}pp=+1 \\ q=-1\end{array}}$$

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

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