CBSE Class 12-science Answered

Evaluate $integral subscript negative 2 end subscript superscript 2 space space space fraction numerator x squared over denominator 1 plus space 5 to the power of x end fraction space space d x$

Asked by ahuja8087 | 15 Mar, 2017, 04:46: PM

Expert Answer

$begin mathsize 16px style straight I equals integral subscript negative 2 end subscript superscript 2 fraction numerator straight x squared over denominator 1 plus 5 to the power of straight x end fraction dx space space space space space...... open parentheses straight i close parentheses straight I equals integral subscript negative 2 end subscript superscript 2 fraction numerator open parentheses negative 2 plus 2 minus straight x close parentheses squared over denominator 1 plus 5 to the power of open parentheses negative 2 plus 2 minus straight x close parentheses end exponent end fraction dx straight I equals integral subscript negative 2 end subscript superscript 2 fraction numerator straight x squared over denominator 1 plus 5 to the power of open parentheses negative straight x close parentheses end exponent end fraction dx straight I equals integral subscript negative 2 end subscript superscript 2 fraction numerator straight x squared over denominator 1 plus begin display style 1 over 5 to the power of straight x end style end fraction dx straight I equals integral subscript negative 2 end subscript superscript 2 fraction numerator 5 to the power of straight x straight x squared over denominator 1 plus begin display style 5 end style begin display style blank to the power of straight x end style end fraction dx space space space space....... open parentheses ii close parentheses Adding space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma 2 straight I equals integral subscript negative 2 end subscript superscript 2 fraction numerator 5 to the power of straight x straight x squared plus straight x squared over denominator 1 plus begin display style 5 end style begin display style blank to the power of straight x end style end fraction dx 2 straight I equals integral subscript negative 2 end subscript superscript 2 fraction numerator open parentheses 5 to the power of straight x plus 1 close parentheses straight x squared over denominator 1 plus begin display style 5 end style begin display style blank to the power of straight x end style end fraction dx 2 straight I equals integral subscript negative 2 end subscript superscript 2 straight x squared dx 2 straight I equals 2 integral subscript 0 superscript 2 straight x squared dx straight I equals open vertical bar straight x cubed over 3 close vertical bar squared subscript 0 straight I equals 2 cubed over 3 straight I equals 8 over 3 end style$

Answered by | 15 Mar, 2017, 05:47: PM