Please solve the question in photo.....

Asked by Vidushi412 | 30th Mar, 2019, 12:50: PM

Expert Answer:

begin mathsize 16px style straight I equals integral fraction numerator open parentheses square root of straight x close parentheses to the power of 5 over denominator open parentheses square root of straight x close parentheses to the power of 7 plus straight x to the power of 6 end fraction dx
straight I equals integral fraction numerator open parentheses square root of straight x close parentheses to the power of 5 over denominator open parentheses square root of straight x close parentheses to the power of 7 plus open parentheses square root of straight x close parentheses to the power of 12 end fraction dx
Put space square root of straight x equals straight t
fraction numerator 1 over denominator 2 square root of straight x end fraction dx equals dt
dx equals 2 tdt
straight I equals integral fraction numerator straight t to the power of 5 over denominator straight t to the power of 7 plus straight t to the power of 12 end fraction 2 tdt
straight I equals integral fraction numerator 2 straight t to the power of 6 dt over denominator straight t to the power of 7 left parenthesis 1 plus straight t to the power of 5 right parenthesis end fraction
straight I equals integral fraction numerator 2 dt over denominator straight t open parentheses 1 plus straight t to the power of 5 close parentheses end fraction
straight I equals integral fraction numerator 2 straight t to the power of 4 dt over denominator straight t to the power of 4 open parentheses 1 plus straight t to the power of 5 close parentheses end fraction
Put space 1 plus straight t to the power of 5 equals straight z space and space then space partial space fraction space you space will space get space the space value space of space straight a space and space straight lambda
where space straight a space plus space straight lambda greater than 2 end style

Answered by Sneha shidid | 30th Mar, 2019, 02:19: PM

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