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Integrals

Question
CBSEENMA12032464

Evaluate the following definite integral:
integral subscript 1 superscript 2 fraction numerator 1 over denominator straight x left parenthesis 1 plus log space straight x right parenthesis squared end fraction dx




Solution

Let I = integral subscript 1 superscript 2 fraction numerator 1 over denominator straight x left parenthesis 1 plus log space straight x right parenthesis squared end fraction dx
Put log x = y;    therefore space space 1 over straight x space dx space equals space dy
When x = 1,    y = log 1 = 0
When x = 2,    y = log 2
therefore     I = integral subscript 0 superscript log space 2 end superscript fraction numerator 1 over denominator left parenthesis 1 plus straight y right parenthesis squared end fraction dy space equals space integral subscript 0 superscript log space 2 end superscript left parenthesis 1 plus straight y right parenthesis to the power of negative 2 end exponent space dy space equals space open square brackets fraction numerator left parenthesis 1 plus straight y right parenthesis to the power of negative 1 end exponent over denominator negative 1 end fraction close square brackets subscript 0 superscript log space 2 end superscript
            equals negative open square brackets fraction numerator 1 over denominator 1 plus straight y end fraction close square brackets subscript 0 superscript log 2 end superscript space equals space minus open parentheses fraction numerator 1 over denominator 1 plus log 2 end fraction minus 1 close parentheses space equals space 1 minus fraction numerator 1 over denominator 1 plus log 2 end fraction equals space fraction numerator log space 2 over denominator 1 plus log 2 end fraction

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