If ${\displaystyle \int \frac{x^2-8x+7}{{\left(x^2-3x-10\right)}^2}dx=P{\log }_e\left|x-5\right|+Q.\frac{1}{x-5}-R.{\log }_e\left|x+2\right|-s.\frac{1}{x+2}+c}$ , then

$\frac{\left(x-1\right)\left(x-7\right)}{{\left(x-5\right)}^2{\left(x+2\right)}^2}=\frac{A}{x-5}+\frac{B}{{\left(x-5\right)}^2}+\frac{C}{x+2}+\frac{D}{{\left(x+2\right)}^2}$

$\left(x-1\right)\left(x-7\right)=A\left(x-5\right){\left(x+2\right)}^2+B{\left(x+2\right)}^2+C\left(x+2\right){\left(x-5\right)}^2+D{\left(x-5\right)}^2$

Put $x=5,B=\frac{-8}{49}$

Put $x=-2$

$D=\frac{27}{49}$

Coefficient of $x^3\Rightarrow A+C=0$ ,

coefficient $x^2$ : $-A+B-8C+D=1$

            $-A-8C=1+\frac{8}{49}-\frac{27}{49}$           

                        $=\frac{49+8-27}{49}=\frac{57-27}{49}=\frac{30}{49}$

            $A+8C=-\frac{30}{49}$

            $-C+8C=\frac{-30}{49}\Rightarrow 7C=\frac{-30}{49}$

            $C=\frac{-30}{343}$         

$\therefore Q=\frac{8}{49};S=\frac{27}{49}$