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Transcript

Ex 7.4, 1 (3๐‘ฅ^2)/(๐‘ฅ^6 + 1) We need to find โˆซ1โ–’(๐Ÿ‘๐’™^๐Ÿ)/(๐’™^๐Ÿ” + ๐Ÿ) ๐’…๐’™ Let ๐’™^๐Ÿ‘=๐’• Diff both sides w.r.t. x 3๐‘ฅ^2=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐’…๐’™=๐’…๐’•/(๐Ÿ‘๐’™^๐Ÿ ) Thus, our equation becomes โˆซ1โ–’(๐Ÿ‘๐’™^๐Ÿ)/(๐’™^๐Ÿ” + ๐Ÿ) ๐’…๐’™ =โˆซ1โ–’(3๐‘ฅ^2)/((๐‘ฅ^3 )^2 + 1) ๐‘‘๐‘ฅ Putting the value of ๐‘ฅ^3=๐‘ก and ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(3๐‘ฅ^2 ) =โˆซ1โ–’(3๐‘ฅ^2)/(๐‘ก^2 + 1) .๐‘‘๐‘ก/(3๐‘ฅ^2 ) =โˆซ1โ–’๐‘‘๐‘ก/(๐‘ก^2 + 1) =โˆซ1โ–’๐’…๐’•/(๐’•^๐Ÿ + (๐Ÿ)^๐Ÿ ) =1/1 tan^(โˆ’1)โกใ€– ๐‘ก/1 ใ€—+๐ถ It is of form โˆซ1โ–’๐‘‘๐‘ก/(๐‘ฅ^2 + ๐‘Ž^2 ) =1/๐‘Ž ใ€–ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1) ใ€—โกใ€–๐‘ฅ/๐‘Žใ€— +๐ถ โˆด Replacing ๐‘ฅ = ๐‘ก and ๐‘Ž by 1 , we get =tan^(โˆ’1)โกใ€– (๐‘ก)ใ€—+๐ถ =ใ€–ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ) ใ€—โก(๐’™^๐Ÿ‘ )+๐‘ช ("Using" ๐‘ก=๐‘ฅ^3 )

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.