Check sibling questions
Chapter 7 Class 12 Integrals
Concept wise

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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 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.