If $\int \frac{3 e^x-7 e^{-x}}{7 e^x+3 e^{-x}} d x=K x+L \log \left(e^{-2 x}+\frac{7}{3}\right)+C$,then $K+L=$

  • A
    $\frac{-3}{38}$
  • B
    $\frac{21}{38}$
  • C
    $\frac{38}{21}$
  • D
    $\frac{-38}{3}$

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