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Ex 7.5, 21 - Integrate 1 / (ex - 1) [Hing: Put ex = t] - Ex 7.5

Ex 7.5, 21 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.5, 21 - Chapter 7 Class 12 Integrals - Part 3

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Transcript

Ex 7.5, 21 Integrate the function 1/((๐‘’^๐‘ฅ โˆ’ 1) ) [Hint : Put ex = t] Let ๐‘’^๐‘ฅ = ๐‘ก Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘’^๐‘ฅ = ๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ = ๐‘‘๐‘ก/๐‘’^๐‘ฅ Therefore โˆซ1โ–’1/((๐‘’^๐‘ฅ โˆ’ 1) ) ๐‘‘๐‘ฅ = โˆซ1โ–’1/((๐‘ก โˆ’ 1) ) ๐‘‘๐‘ก/๐‘’^๐‘ฅ = โˆซ1โ–’๐‘‘๐‘ก/(๐‘ก(๐‘ก โˆ’ 1) ) We can write integrand as 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 = ๐ดร—(โˆ’1) 1 = โˆ’๐ด ๐ด = โˆ’1 Putting t = 1 1 = ๐ด(1โˆ’1)+๐ตร—1 1 = ๐ดร—0+๐ต 1 = ๐ต ๐ต = 1 Therefore โˆซ1โ–’1/(๐‘ก(๐‘ก โˆ’ 1) ) ๐‘‘๐‘ก = โˆซ1โ–’(โˆ’1)/(๐‘ก ) ๐‘‘๐‘ก + โˆซ1โ–’1/(๐‘ก โˆ’ 1 ) = โˆ’ใ€–log ใ€—โก|๐‘ก|+ใ€–log ใ€—โก|๐‘กโˆ’1|+๐ถ = ใ€–log ใ€—โก|(๐‘ก โˆ’ 1)/๐‘ก|+๐ถ Putting back t = ๐‘’^๐‘ฅ = ใ€–๐‘™๐‘œ๐‘” ใ€—โก|(๐‘’^๐‘ฅ โˆ’ 1)/๐‘’^๐‘ฅ |+๐ถ = ใ€–๐ฅ๐จ๐  ใ€—โก((๐’†^๐’™ โˆ’ ๐Ÿ)/๐’†^๐’™ )+๐‘ช Since ex > 1 for x > 0 โˆด ex โ€“ 1 > 0 โ‡’ |(๐’†^๐’™ โˆ’ ๐Ÿ)/๐’†^๐’™ |=((๐’†^๐’™ โˆ’ ๐Ÿ)/๐’†^๐’™ ) ("As " ๐‘™๐‘œ๐‘” ๐ดโˆ’๐‘™๐‘œ๐‘” ๐ต" = " ๐‘™๐‘œ๐‘” ๐ด/๐ต)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.