If $\frac{e^x + 2}{(e^x - 1)(2e^x - 3)} = -\frac{3}{e^x - 1} + \frac{B}{2e^x - 3}$,then $B = $

  • A
    $1$
  • B
    $3$
  • C
    $5$
  • D
    $7$

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