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Solution of differential equation $\left( {1 + {e^{2y}}} \right){e^{{{\tan }^{ - 1}}x}}dx - \left( {1 + {x^2}} \right)\left( {{e^y} + {{\left( {{e^y} - 1} \right)}^2}} \right)dy = 0$ is

• A
  $\ln \left( y \right) = \tan \left( {y - {e^{{{\tan }^{ - 1}}x}} + C} \right)$
• ✓
  $y = \ln \left( {\tan \left( {y - {e^{{{\tan }^{ - 1}}x}} + C} \right)} \right)$
• C
  $\ln \left( y \right) = \tan \left( {{e^{{{\tan }^{ - 1}}x}} - y + C} \right)$
• D
  $y = \ln \left( {\tan \left( {{e^{{{\tan }^{ - 1}}x}} - y + C} \right)} \right)$