If $\frac{32x^2+186x}{(x^2+1)(x+5)}=\frac{37x+1}{x^2+1}+\frac{\lambda}{x+5}$,then $\frac{\lambda}{2}$ is equal to

  • A
    $-5$
  • B
    $\frac{-7}{2}$
  • C
    $\frac{-3}{2}$
  • D
    $\frac{-5}{2}$

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