Ex 7.4, 8 - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 7.4, 8 Integrate ๐ฅ^2/โ(๐ฅ^6 + ๐^6 ) Let ๐ฅ^3=๐ก Differentiating both sides w.r.t. x 3๐ฅ^2=๐๐ก/๐๐ฅ ๐๐ฅ=๐๐ก/(3๐ฅ^2 ) Integrating the function โซ1โ๐ฅ^2/โ(๐ฅ^6 + ๐^6 ) ๐๐ฅ=โซ1โ๐ฅ^2/โ((๐ฅ^3 )^2 + (๐^3 )^2 ) ๐๐ฅ Putting values of ๐ฅ^3=๐ก and ๐๐ฅ=๐๐ก/(3๐ฅ^2 ) , we get =โซ1โ๐ฅ^2/โ(๐ก^2 + (๐^3 )^2 ) ๐๐ฅ =โซ1โ๐ฅ^2/โ(๐ก^2 + (๐^3 )^2 ) . ๐๐ก/(3๐ฅ^2 ) =โซ1โ1/โ((๐ก^2 + (๐^3 )^2 ) ) . ๐๐ก/3 =1/3 โซ1โ๐๐ก/โ(๐ก^2 + (๐^3 )^2 ) =1/3 [logโก|๐ก+โ(๐ก^2 + (๐^3 )^2 )|+๐ถ1] It is of form โซ1โ๐๐ฅ/โ(๐ฅ^2 + ๐^2 ) =logโก|๐ฅ+โ(๐ฅ^2 + ๐^2 )|+๐ถ1 โด Replacing ๐ฅ by ๐ก and a by ๐^3, we get =1/3 logโก|๐ก+โ(๐ก^2 + ๐^6 ) |+๐ถ =1/3 logโก|๐ฅ^3+โ((๐ฅ^3 )^2 + ๐^6 ) |+๐ถ =๐/๐ ๐๐๐โก|๐^๐+โ(๐^๐+ ๐^๐ ) |+๐ช ("Using" ๐ก=๐ฅ^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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo