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

Transcript

Ex 7.4, 8 Integrate ๐‘ฅ^2/โˆš(๐‘ฅ^6 + ๐‘Ž^6 ) Let ๐‘ฅ^3=๐‘ก Differentiating both sides w.r.t. x 3๐‘ฅ^2=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(3๐‘ฅ^2 ) Integrating the function โˆซ1โ–’๐‘ฅ^2/โˆš(๐‘ฅ^6 + ๐‘Ž^6 ) ๐‘‘๐‘ฅ=โˆซ1โ–’๐‘ฅ^2/โˆš((๐‘ฅ^3 )^2 + (๐‘Ž^3 )^2 ) ๐‘‘๐‘ฅ Putting values of ๐‘ฅ^3=๐‘ก and ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(3๐‘ฅ^2 ) , we get =โˆซ1โ–’๐‘ฅ^2/โˆš(๐‘ก^2 + (๐‘Ž^3 )^2 ) ๐‘‘๐‘ฅ =โˆซ1โ–’๐‘ฅ^2/โˆš(๐‘ก^2 + (๐‘Ž^3 )^2 ) . ๐‘‘๐‘ก/(3๐‘ฅ^2 ) =โˆซ1โ–’1/โˆš((๐‘ก^2 + (๐‘Ž^3 )^2 ) ) . ๐‘‘๐‘ก/3 =1/3 โˆซ1โ–’๐‘‘๐‘ก/โˆš(๐‘ก^2 + (๐‘Ž^3 )^2 ) =1/3 [logโก|๐‘ก+โˆš(๐‘ก^2 + (๐‘Ž^3 )^2 )|+๐ถ1] It is of form โˆซ1โ–’๐‘‘๐‘ฅ/โˆš(๐‘ฅ^2 + ๐‘Ž^2 ) =logโก|๐‘ฅ+โˆš(๐‘ฅ^2 + ๐‘Ž^2 )|+๐ถ1 โˆด Replacing ๐‘ฅ by ๐‘ก and a by ๐‘Ž^3, we get =1/3 logโก|๐‘ก+โˆš(๐‘ก^2 + ๐‘Ž^6 ) |+๐ถ =1/3 logโก|๐‘ฅ^3+โˆš((๐‘ฅ^3 )^2 + ๐‘Ž^6 ) |+๐ถ =๐Ÿ/๐Ÿ‘ ๐’๐’๐’ˆโก|๐’™^๐Ÿ‘+โˆš(๐’™^๐Ÿ”+ 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 9 years. He provides courses for Maths and Science at Teachoo.