If $\int \frac{1-x^9}{x\left(1+x^9\right)}dx=A\log \left|x\right|+B\log \left|1+x^9\right|+C,$ then the value of $\frac{A}{B}$ is equal to

$\int \frac{1-x^9}{x\left(1+x^9\right)}dx=A\log \left|x\right|+B\log \left|1+x^9\right|+C$
Differentiating both the sides, we get
$\begin{array}{l}\frac{1-x^9}{x\left(1+x^9\right)}=\frac{A}{x}+\frac{9B{x}^8}{1+x^9}\\ \Rightarrow 1-x^9=A\left(1+x^9\right)+9B{x}^9\\ \Rightarrow A=1,-1=A+9B\\ \Rightarrow A=1,B=-\frac{2}{9}\end{array}$
Thus, $\frac{A}{B}=-\frac{9}{2}=-4.5$