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

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)

About the Author

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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.