If $\frac{\sin^2 x + 1}{2\sin^2 x - 5\sin x + 3} = \frac{A}{2\sin x - 3} + \frac{B}{\sin x - 1} + C$,then:

  • A
    $A = \frac{13}{2}$
  • B
    $B = 2$
  • C
    $C = 1$
  • D
    $A + B + C = 5$

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