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Please solve the following question

Asked by Balbir 7th October 2017, 10:59 PM
Answered by Expert
Answer:
begin mathsize 16px style integral square root of fraction numerator straight a plus straight x over denominator straight x end fraction end root dx
integral square root of fraction numerator left parenthesis straight a plus straight x right parenthesis left parenthesis straight a plus straight x right parenthesis over denominator straight x left parenthesis straight a plus straight x right parenthesis end fraction end root dx
integral fraction numerator straight a plus straight x over denominator square root of straight x left parenthesis straight a plus straight x right parenthesis end root end fraction dx
Consider comma
straight a plus straight x space equals space straight A straight d over dx open parentheses ax plus straight x squared close parentheses plus straight B
straight a plus straight x equals straight A left parenthesis straight a plus 2 straight x right parenthesis plus straight B........... left parenthesis 1 right parenthesis
straight a plus straight x equals Aa plus 2 Ax plus straight B
straight a plus straight x equals Aa plus straight B plus 2 Ax
Comparing space on space both space sides comma
2 straight A equals 1 rightwards double arrow straight A equals 1 half
Put space it space in space Aa plus straight B equals straight a rightwards double arrow straight B equals straight a over 2
integral fraction numerator begin display style 1 half end style left parenthesis straight a plus 2 straight x right parenthesis plus begin display style straight a over 2 end style over denominator square root of ax plus straight x squared end root end fraction dx
Separate space the space terms space and space integrate space further.

Put space it space in space equation space left parenthesis 1 right parenthesis end style
Answered by Expert 22nd December 2017, 9:31 AM
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