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1. Chapter 7 Class 12 Integrals
2. Concept wise
3. Integration by specific formulaes - Formula 6

Transcript

Ex 7.4, 8 Integrate ๐ฅ^2/โ(๐ฅ^6 + ๐^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 โซ1โ๐ฅ^2/โ(๐ฅ^6 + ๐^6 ) ๐๐ฅ=โซ1โ๐ฅ^2/โ((๐ฅ^3 )^2 + (๐^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 =โซ1โ๐ฅ^2/โ(๐ก^2 + (๐^3 )^2 ) ๐๐ฅ =โซ1โ๐ฅ^2/โ(๐ก^2 + (๐^3 )^2 ) . ๐๐ก/(3๐ฅ^2 ) =โซ1โ1/โ((๐ก^2 + (๐^3 )^2 ) ) . ๐๐ก/3 =1/3 โซ1โ๐๐ก/โ(๐ก^2 + (๐^3 )^2 ) =1/3 [logโก|๐ก+โ(๐ก^2 + (๐^3 )^2 )|+๐ถ1] It is of form โซ1โ๐๐ฅ/โ(๐ฅ^2 + ๐^2 ) =logโก|๐ฅ+โ(๐ฅ^2 + ๐^2 )|+๐ถ1 โด Replacing ๐ฅ by ๐ก and a by ๐^3, we get =1/3 logโก|๐ก+โ(๐ก^2 + ๐^6 ) |+๐ถ =1/3 logโก|๐ฅ^3+โ((๐ฅ^3 )^2 + ๐^6 ) |+๐ถ =๐/๐ ๐๐๐โก|๐^๐+โ(๐^๐+ ๐^๐ ) |+๐ช ("Using" ๐ก=๐ฅ^3 )

Integration by specific formulaes - Formula 6