Chapter 7 Class 12 Integrals

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Ex 7.4, 8 - Chapter 7 Class 12 Integrals (Important Question)

Last updated at July 14, 2026 by

Ex 7.4, 8 - Integrate x2 / root x6 + a6 - Chapter 7 NCERT

Ex 7.4, 8 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.4, 8 - Chapter 7 Class 12 Integrals - Part 3

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Ex 7.4, 8 Integrate š‘„^2/āˆš(š‘„^6 + š‘Ž^6 ) Let š‘„^3=š‘” Differentiating both sides w.r.t. x 3š‘„^2=š‘‘š‘”/š‘‘š‘„ š‘‘š‘„=š‘‘š‘”/(3š‘„^2 ) Integrating the function āˆ«1ā–’š‘„^2/āˆš(š‘„^6 + š‘Ž^6 ) š‘‘š‘„=āˆ«1ā–’š‘„^2/āˆš((š‘„^3 )^2 + (š‘Ž^3 )^2 ) š‘‘š‘„ Putting values of š‘„^3=š‘” and š‘‘š‘„=š‘‘š‘”/(3š‘„^2 ) , we get =āˆ«1ā–’š‘„^2/āˆš(š‘”^2 + (š‘Ž^3 )^2 ) š‘‘š‘„ =āˆ«1ā–’š‘„^2/āˆš(š‘”^2 + (š‘Ž^3 )^2 ) . š‘‘š‘”/(3š‘„^2 ) =āˆ«1ā–’1/āˆš((š‘”^2 + (š‘Ž^3 )^2 ) ) . š‘‘š‘”/3 =1/3 āˆ«1ā–’š‘‘š‘”/āˆš(š‘”^2 + (š‘Ž^3 )^2 ) =1/3 [logā”|š‘”+āˆš(š‘”^2 + (š‘Ž^3 )^2 )|+š¶1] It is of form āˆ«1ā–’š‘‘š‘„/āˆš(š‘„^2 + š‘Ž^2 ) =logā”|š‘„+āˆš(š‘„^2 + š‘Ž^2 )|+š¶1 āˆ“ Replacing š‘„ by š‘” and a by š‘Ž^3, we get =1/3 logā”|š‘”+āˆš(š‘”^2 + š‘Ž^6 ) |+š¶ =1/3 logā”|š‘„^3+āˆš((š‘„^3 )^2 + š‘Ž^6 ) |+š¶ =šŸ/šŸ‘ š’š’š’ˆā”|š’™^šŸ‘+āˆš(š’™^šŸ”+ š’‚^šŸ” ) |+š‘Ŗ ("Using" š‘”=š‘„^3 )

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