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CBSE Class 12-science Answered

Please solve the following question

Asked by Balbir | 07 Oct, 2017, 11:21: PM
answered-by-expert Expert Answer
begin mathsize 16px style integral square root of straight x to the power of 2 straight n end exponent space plus space 1 end root space cross times space fraction numerator log open parentheses straight x to the power of 2 straight n end exponent plus 1 close parentheses minus 2 nlogx over denominator straight x to the power of 3 straight n plus 1 end exponent end fraction dx integral fraction numerator square root of straight x to the power of 2 straight n end exponent end root open parentheses 1 plus begin display style 1 over straight x to the power of 2 straight n end exponent end style close parentheses to the power of begin display style 1 half end style end exponent over denominator straight x to the power of 3 straight n plus 1 end exponent end fraction cross times open square brackets log space open parentheses straight x to the power of 2 straight n end exponent plus 1 close parentheses minus log space straight x to the power of 2 straight n end exponent close square brackets dx integral fraction numerator straight x to the power of straight n open parentheses 1 plus 1 over straight x to the power of 2 straight n end exponent close parentheses to the power of 1 half end exponent over denominator straight x to the power of 3 straight n plus 1 end exponent end fraction cross times log open parentheses fraction numerator straight x to the power of 2 straight n end exponent plus 1 over denominator space straight x to the power of 2 straight n end exponent end fraction close parentheses dx integral fraction numerator straight x to the power of straight n open parentheses 1 plus 1 over straight x to the power of 2 straight n end exponent close parentheses to the power of 1 half end exponent over denominator straight x to the power of 3 straight n plus 1 end exponent end fraction cross times log open parentheses 1 plus 1 over straight x to the power of 2 straight n end exponent close parentheses dx integral open parentheses 1 plus 1 over straight x to the power of 2 straight n end exponent close parentheses to the power of 1 half end exponent over straight x to the power of 2 straight n plus 1 end exponent cross times log open parentheses 1 plus 1 over straight x to the power of 2 straight n end exponent close parentheses dx 1 plus 1 over straight x to the power of 2 straight n end exponent equals straight t rightwards double arrow negative 2 nx to the power of negative 2 straight n minus 1 end exponent dx equals dt fraction numerator negative 2 straight n over denominator straight x to the power of 2 straight n plus 1 end exponent end fraction dx equals dt 1 over straight x to the power of 2 straight n plus 1 end exponent dx equals fraction numerator dt over denominator negative 2 straight n end fraction integral square root of straight t space log space straight t space cross times open parentheses fraction numerator dt over denominator negative 2 straight n end fraction close parentheses Solve space taking space straight u equals space log space straight t space and space straight v space equals space square root of straight t space and space using space parttial space integration space formul. end style
Answered by Sneha shidid | 21 Dec, 2017, 02:58: PM

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