If $\frac{(x + 1)^2}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$,then $\sin^{-1}\left(\frac{A}{C}\right) = $

  • A
    $\frac{\pi}{6}$
  • B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{3}$
  • D
    $\frac{\pi}{2}$

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