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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 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.