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Ex 7.2, 19 - Integrate e2x - 1 / e2x + 1 - Chapter 7 - Ex 7.2

Ex 7.2, 19 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.2, 19 - Chapter 7 Class 12 Integrals - Part 3

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Ex 7.2, 19 Integrate the function (๐‘’2๐‘ฅ โˆ’ 1)/(๐‘’2๐‘ฅ+ 1) Simplify the given function (๐‘’^2๐‘ฅ โˆ’ 1)/(๐‘’^2๐‘ฅ + 1) Dividing numerator and denominator by ex, we obtain = (๐‘’^2๐‘ฅ/๐‘’^๐‘ฅ " " โˆ’" " ๐Ÿ/๐’†^๐’™ )/(๐‘’^2๐‘ฅ/๐‘’^๐‘ฅ " " + " " ๐Ÿ/๐’†^๐’™ ) = (๐‘’^๐’™ โˆ’ ๐’†^(โˆ’๐’™))/(๐‘’^๐’™ + ๐’†^(โˆ’๐’™) ) Let ๐‘’^๐‘ฅ + ๐‘’^(โˆ’๐‘ฅ)= ๐‘ก Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘’^๐‘ฅ+(โˆ’1) ๐‘’^(โˆ’๐‘ฅ)= ๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘’^๐‘ฅโˆ’๐‘’^(โˆ’๐‘ฅ)= ๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(๐‘’^๐‘ฅ โˆ’ ๐‘’^(โˆ’๐‘ฅ) ) Now, Integrating the function โˆซ1โ–’ใ€–" " (๐‘’^2๐‘ฅ โˆ’ 1)/(๐‘’^2๐‘ฅ + 1) " " ใ€—. ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–" " (๐‘’^๐‘ฅ โˆ’ ๐‘’^(โˆ’๐‘ฅ))/(๐‘’^๐‘ฅ + ๐‘’^(โˆ’๐‘ฅ) ) " " ใ€—. ๐‘‘๐‘ฅ (Using (๐‘’^2๐‘ฅ โˆ’ 1)/(๐‘’^2๐‘ฅ + 1)=(๐‘’^๐‘ฅ โˆ’ ๐‘’^(โˆ’๐‘ฅ))/(๐‘’^๐‘ฅ + ๐‘’^(โˆ’๐‘ฅ) ) ) Putting ๐‘’^๐‘ฅ + ๐‘’^(โˆ’๐‘ฅ)=๐‘ก & ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(๐‘’^๐‘ฅ โˆ’ ๐‘’^(โˆ’๐‘ฅ) ) =โˆซ1โ–’ใ€–" " (๐‘’^๐‘ฅ โˆ’ ๐‘’^(โˆ’๐‘ฅ))/๐‘ก " " ใ€—. ๐‘‘๐‘ก/(๐‘’^๐‘ฅ โˆ’ ๐‘’^(โˆ’๐‘ฅ) ) " " =โˆซ1โ–’ใ€–" " 1/๐‘ก " " ใ€—. ๐‘‘๐‘ก =logโก|๐‘ก|+๐ถ =logโกใ€– |๐‘’^๐‘ฅ+๐‘’^(โˆ’๐‘ฅ) |ใ€—+๐ถ =๐’๐’๐’ˆโก(๐’†^๐’™ + ๐’†^(โˆ’๐’™) )+๐‘ช (Using ๐‘ก=๐‘’^๐‘ฅ + ๐‘’^(โˆ’๐‘ฅ)) (As ๐‘’^๐‘ฅ+๐‘’^(โˆ’๐‘ฅ)>0 )

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.