Please solve the following question

Asked by Balbir | 7th Oct, 2017, 10:59: PM

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 Sneha shidid | 22nd Dec, 2017, 09:31: AM