1. Chapter 7 Class 12 Integrals
  2. Serial order wise
  3. Ex 7.4
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Ex 7.4, 1 - Chapter 7 Class 12 Integrals

Last updated at Dec. 16, 2024 by Teachoo

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

    3. Ex 7.4

    Ex 7.5 →
Ex 7.5 →
Chapter 7 Class 12 Integrals
Serial order wise

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo