If  $\frac{2x+x^2-1}{x(x^2-1)}=\frac{A}{x}+\frac{B}{x-1}-\frac{C}{x+1},\theta \in Q_1$ then   ${\sin }^{-1}\left(\frac{A}{B}\right)=$

$\frac{2x+x^2-1}{x(x^2-1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$

$\Rightarrow 2x+x^2-1=A(x^2-1)+B\left(x\right)(x+1)+C\left(x\right)(x-1)$

By comparing equal coefficients both sides we get  $A=1,B=1,C=-1$

$\therefore {\sin }^{-1}\left(\frac{A}{B}\right)={\sin }^{-1}\left(1\right)=\frac{\pi }{2}$