Question

Evaluate $integral subscript negative 2 end subscript superscript 2 space log space open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses space dx.$

Name: Wired Faculty - The Best Learning App
Rating: 4.6 (1864 reviews)

Solution

integral subscript negative straight a end subscript superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space minus integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx plus integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx

rightwards double arrow

integral subscript negative straight a end subscript superscript straight a space straight f left parenthesis straight x right parenthesis space dx space equals space 0

Let I =

integral subscript negative 2 end subscript superscript 2 space log space open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses dx

Let

straight f left parenthesis straight x right parenthesis space equals space log space open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses

therefore space space space space space straight f left parenthesis negative straight x right parenthesis space equals space log open parentheses fraction numerator 2 minus straight x over denominator 2 plus straight x end fraction close parentheses space equals space log open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses to the power of negative 1 end exponent space equals space minus log open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses space equals space minus straight f left parenthesis straight x right parenthesis

therefore space space space space straight f left parenthesis straight x right parenthesis space is space an space odd space function space of space straight x

therefore space space space space space straight I space equals space 0 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 space space integral subscript negative straight a end subscript superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space 0 space if space straight f left parenthesis straight x right parenthesis space is space an space odd space function space of space straight x close square brackets

For Daily Free Study Material Join wiredfaculty