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Integrals

Question
CBSEENMA12032396
Wired Faculty App

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

Solution

Let I = integral subscript 1 superscript 2 fraction numerator 1 over denominator left parenthesis straight x plus 1 right parenthesis space left parenthesis straight x plus 2 right parenthesis end fraction dx space equals space integral subscript 1 superscript 2 open square brackets fraction numerator 1 over denominator left parenthesis straight x plus 1 right parenthesis left parenthesis negative 1 plus 2 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis negative 2 plus 1 right parenthesis space left parenthesis straight x plus 2 end fraction close square brackets dx
         equals space integral subscript 1 superscript 2 open parentheses fraction numerator 1 over denominator straight x plus 1 end fraction minus fraction numerator 1 over denominator straight x plus 2 end fraction close parentheses dx space equals space open square brackets log left parenthesis straight x plus 1 right parenthesis minus log left parenthesis straight x plus 2 right parenthesis close square brackets subscript 1 superscript 2 equals space open square brackets log open parentheses fraction numerator straight x plus 1 over denominator straight x plus 2 end fraction close parentheses close square brackets subscript 1 superscript 2 space equals space log 3 over 4 minus log 2 over 3 equals space log open parentheses fraction numerator begin display style 3 over 4 end style over denominator begin display style 2 over 3 end style end fraction close parentheses space equals space log open parentheses 9 over 8 close parentheses

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