If $\frac{x^2}{\left(x^2+2\right)\left(x^4-1\right)}=\frac{A}{x^2-1}+\frac{B}{x^2+1}+\frac{C}{x^2+2}$, then $A+B-C=$

• A

  $0$
• Correct answer
  B

  $\frac{4}{3}$
• C

  $\frac{3}{4}$
• D

  $2$

Hence, the correct answer is option: 'B'.