Integration by specific formulaes - Formula 7

Chapter 7 Class 12 Integrals
Concept wise

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Ex 7.7, 11 β«1βγβ(π₯2 β8π₯+7) " " γ ππ₯ 1/2 (x - 4)β(π₯2 β8π₯+7) + 9 log |π₯β4+β(π₯2β8π₯+7)| + C 1/2 (x + 4)β(π₯2 β8π₯+7) + 9 log |π₯+4+β(π₯2β8π₯+7)| + C 1/2 (x - 4)β(π₯2 β8π₯+7) - 3β2 log |π₯β4+β(π₯2β8π₯+7)| + C 1/2 (x - 4)β(π₯2 β8π₯+7) + 9/2 log |π₯β4+β(π₯2β8π₯+7)| + C β«1βγβ(π₯^2β8π₯+7) ππ₯γ =β«1βγβ(π₯^2β2(4)(π₯)+7) ππ₯γ =β«1βγβ(π₯^2β2(4)(π₯) γ+(4)γ^2β(4)^2+7) ππ₯γ =β«1βγβ((π₯β4)^2β16+7) ππ₯γ =β«1βγβ((π₯β4)^2β9 ) ππ₯γ =β«1βγβ((π₯β4)^2β(3)^2 ) ππ₯γ =(π₯ β 4)/2 β((π₯β4)^2β(3)^2 )β(3)^2/2 πππ|π₯β4+β((π₯β4)^2β(3)^2 )|+πΆ =(π₯ β 4)/2 β(π₯^2β8π₯+16β9 )β9/2 πππ|π₯β4+β(π₯^2β8π₯+16β9)|+πΆ =(π₯ β 4)/2 β(π₯^2β8π₯+7)β9/2 πππ|π₯β4+β(π₯^2β8π₯+7)|+πΆ β΄ Option (D) is correct. It is of the form β«1βγβ(π₯^2βπ^2 ) ππ₯=π₯/2 β(π₯^2βπ^2 )βπ^2/2 πππ|π₯+β(π₯^2βπ^2 )|+πΆγ β΄ Replacing π₯ by π₯β4 and a by 3 , we get