Integration by specific formulaes - Formula 3

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 โซ1โ(๐๐^๐)/(๐^๐ + ๐) ๐๐ Let ๐^๐=๐ Diff both sides w.r.t. x 3๐ฅ^2=๐๐ก/๐๐ฅ ๐๐=๐๐/(๐๐^๐ ) Thus, our equation becomes โซ1โ(๐๐^๐)/(๐^๐ + ๐) ๐๐ =โซ1โ(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 ) =โซ1โ(3๐ฅ^2)/(๐ก^2 + 1) .๐๐ก/(3๐ฅ^2 ) =โซ1โ๐๐ก/(๐ก^2 + 1) =โซ1โ๐๐/(๐^๐ + (๐)^๐ ) =1/1 tan^(โ1)โกใ ๐ก/1 ใ+๐ถ It is of form โซ1โ๐๐ก/(๐ฅ^2 + ๐^2 ) =1/๐ ใใ๐ก๐๐ใ^(โ1) ใโกใ๐ฅ/๐ใ +๐ถ โด Replacing ๐ฅ = ๐ก and ๐ by 1 , we get =tan^(โ1)โกใ (๐ก)ใ+๐ถ =ใใ๐๐๐ใ^(โ๐) ใโก(๐^๐ )+๐ช ("Using" ๐ก=๐ฅ^3 )