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The indefinite integral of left parenthesis 12 s i n invisible function application x plus 5 c o s invisible function application x right parenthesis to the power of negative 1 end exponent is, for any arbitrary constant -

  1. 1 divided by 13 log invisible function application open vertical bar tan invisible function application open parentheses x over 2 plus 1 half tan to the power of negative 1 end exponent invisible function application 5 over 12 close parentheses close vertical bar plus c    
  2. 1 divided by 13 l o g invisible function application open vertical bar t a n invisible function application open parentheses fraction numerator x over denominator 2 end fraction plus fraction numerator 1 over denominator 2 end fraction t a n to the power of negative 1 end exponent invisible function application fraction numerator 12 over denominator 5 end fraction close parentheses close vertical bar plus c    
  3. 1 divided by 13 l o g invisible function application open vertical bar t a n invisible function application open parentheses x plus t a n to the power of negative 1 end exponent invisible function application fraction numerator 5 over denominator 12 end fraction close parentheses close vertical bar plus c    
  4. 1 divided by 13 log invisible function application open vertical bar tan invisible function application open parentheses x plus tan to the power of negative 1 end exponent invisible function application 12 over 5 close parentheses close vertical bar plus c  

The correct answer is: 1 divided by 13 l o g invisible function application open vertical bar t a n invisible function application open parentheses fraction numerator x over denominator 2 end fraction plus fraction numerator 1 over denominator 2 end fraction t a n to the power of negative 1 end exponent invisible function application fraction numerator 12 over denominator 5 end fraction close parentheses close vertical bar plus c



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