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Integrals

Question
CBSEENMA12036188

let space straight F left parenthesis straight x right parenthesis space equals space straight f left parenthesis straight x right parenthesis space plus space straight f space open parentheses 1 over straight x close parentheses comma space where space straight f left parenthesis straight x right parenthesis space equals space integral subscript 1 superscript straight x fraction numerator log space straight t over denominator 1 plus straight t end fraction dt. Then F(e) equals
  • 1/2

  • 1

  • 2

  • 0

Solution

A.

1/2

straight f left parenthesis straight x right parenthesis space equals space integral subscript 1 superscript straight x fraction numerator log space straight t over denominator 1 space plus straight t end fraction space dt
straight F left parenthesis straight e right parenthesis space equals space straight f left parenthesis straight e right parenthesis space plus space straight f open parentheses 1 over straight e close parentheses
straight F left parenthesis straight e right parenthesis space equals space integral subscript 1 superscript straight e space fraction numerator log space straight t over denominator 1 plus straight t end fraction dt space plus integral subscript 1 superscript 1 divided by straight e end superscript fraction numerator log space straight t over denominator 1 plus straight t end fraction dt
space equals space integral subscript 1 superscript straight e fraction numerator log space straight t over denominator 1 plus straight t end fraction space plus integral subscript 1 superscript straight e fraction numerator log space straight t over denominator straight t left parenthesis 1 plus straight t right parenthesis end fraction dt
space equals space integral subscript 1 superscript straight e space fraction numerator log space straight t over denominator straight t end fraction space dt space equals space 1 half

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