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Integrals

Question
CBSEENMA12032668
Wired Faculty App

The value of integral subscript 0 superscript 1 space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x minus 1 over denominator 1 plus straight x minus straight x squared end fraction close parentheses dx is 

  • 1

  • 0

  • -1

  • straight pi over 4

Solution

B.

0

Let I = integral subscript 0 superscript 1 tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x minus 1 over denominator 1 plus straight x minus straight x squared end fraction close parentheses dx
         equals space integral subscript 0 superscript 1 tan to the power of negative 1 end exponent open curly brackets fraction numerator straight x plus left parenthesis straight x minus 1 right parenthesis over denominator 1 minus straight x left parenthesis straight x minus 1 right parenthesis end fraction close curly brackets space dx space equals space integral subscript 0 superscript 1 left square bracket tan to the power of negative 1 end exponent straight x space plus space tan to the power of negative 1 end exponent left parenthesis straight x minus 1 right parenthesis right square bracket space dx
therefore space space space space space straight I space equals integral subscript 0 superscript 1 space open square brackets tan to the power of negative 1 end exponent minus space tan to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis close square brackets dx                         ...(1)
Again, straight I space equals space integral subscript 0 superscript 1 open square brackets tan to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis space minus space tan to the power of negative 1 end exponent open curly brackets 1 minus left parenthesis 1 minus straight x close curly brackets close square brackets space dx
                                                              open square brackets because space space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript straight a superscript straight b straight f left parenthesis straight a plus straight b minus straight x right parenthesis space dx close square brackets
                 equals space integral subscript 0 superscript 1 left square bracket tan to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis space minus space tan to the power of negative 1 end exponent straight x right square bracket dx space equals space integral subscript 0 superscript 1 left square bracket tan to the power of negative 1 end exponent straight x space minus space tan to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis right square bracket space dx
therefore space space space space space space straight I space equals space minus straight I space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
therefore space space space space space space 2 space space straight I space equals space 0 space space space space space space space space space space space rightwards double arrow space space space space space straight I space equals space 0 therefore space space space integral subscript 0 superscript 1 tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x minus 1 over denominator 1 plus straight x minus straight x squared end fraction close parentheses dx space equals space 0
therefore space space space left parenthesis straight B right parenthesis space is space correct space answer.

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