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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Ex 7.5, 23 โˆซ1โ–’๐‘‘๐‘ฅ/(๐‘ฅ(๐‘ฅ2 + 1) ) equals log |๐‘ฅ| โ€“ 1/(2 ) log (x2+1) + C log |๐‘ฅ| + 1/(2 ) log (x2+1) + C log |๐‘ฅ| + 1/(2 ) log (x2+1) + C 1/(2 ) log |๐‘ฅ| + log (x2+1) + C โˆซ1โ–’๐‘‘๐‘ฅ/(๐‘ฅ(๐‘ฅ2 + 1) ) Let t = (1+๐‘ฅ^2) Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๐‘ก/๐‘‘๐‘ฅ = 0+2๐‘ฅ ๐‘‘๐‘ฅ = ๐‘‘๐‘ก/2๐‘ฅ Thus, our equation becomes โˆซ1โ–’๐‘‘๐‘ฅ/(๐‘ฅ(๐‘ฅ^2 + 1) ) = โˆซ1โ–’1/(๐‘ฅ ๐‘ก) ๐‘‘๐‘ก/2๐‘ฅ = 1/2 โˆซ1โ–’ใ€– ๐‘‘๐‘ก/๐‘กร—1/๐‘ฅ^2 ใ€— = 1/2 โˆซ1โ–’๐‘‘๐‘ก/(๐‘ก(๐‘ก โˆ’ 1)) We can write integrand as Since t = (1+๐‘ฅ^2) x2 = (t โ€“ 1) 1/(๐‘ก(๐‘ก โˆ’ 1)) = ๐ด/๐‘ก + ๐ต/(๐‘ก โˆ’ 1) 1/(๐‘ก(๐‘ก โˆ’ 1)) = (๐ด(๐‘ก โˆ’ 1) + ๐ต๐‘ก)/(๐‘ก(๐‘ก โˆ’ 1)) Cancelling denominator 1 = ๐ด(๐‘กโˆ’1)+๐ต๐‘ก Putting t = 0 in (1) 1 = ๐ด(0โˆ’1)+๐ตร—0 1 = โˆ’๐ด ๐ด = โˆ’1 Putting t = 1 in (1) 1 = ๐ด(1โˆ’1)+๐ต(1) 1 = ๐ดร—0+๐ต 1 = ๐ต ๐ต = 1 Therefore 1/2 โˆซ1โ–’๐‘‘๐‘ก/(๐‘ก(๐‘ก โˆ’ 1)) = 1/2 โˆซ1โ–’ใ€–(โˆ’1)/๐‘ก ใ€— ๐‘‘๐‘ก + 1/2 โˆซ1โ–’ใ€–1/(๐‘ก โˆ’ 1) ใ€— ๐‘‘๐‘ก = (โˆ’1)/2 ใ€–log ใ€—โก|๐‘ก|+ 1/2 ใ€–log ใ€—โก|๐‘กโˆ’1|+๐ถ Putting back t = (1+๐‘ฅ^2) = (โˆ’1)/2 ใ€–log ใ€—โก|1+๐‘ฅ^2 |+ 1/2 ใ€–log ใ€—โก|1+๐‘ฅ^2โˆ’1|+๐ถ = (โˆ’1)/2 ใ€–log ใ€—โก|1+๐‘ฅ^2 |+ 1/2 ใ€–log ใ€—โก|๐‘ฅ^2 |+๐ถ = (โˆ’1)/2 ใ€–log ใ€—โก|1+๐‘ฅ^2 |+ 1/2ร—2 ใ€–log ใ€—โก|๐‘ฅ|+๐ถ = (โˆ’1)/2 ใ€–log ใ€—โก|1+๐‘ฅ^2 |+ ใ€–log ใ€—โก|๐‘ฅ|+๐ถ As 1 + ๐‘ฅ^2 it is always positive = ใ€–log ใ€—โก|๐‘ฅ|โˆ’ 1/2 ใ€–log ใ€—โกใ€–(๐‘ฅ^2+1)ใ€—+๐ถ โˆด Correct answer is A . (log ๐‘ฅ^๐‘Ž=๐‘Ž logโก๐‘ฅ)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.