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Ex 7.7

Ex 7.7, 1

Ex 7.7, 2 Important

Ex 7.7, 3

Ex 7.7, 4

Ex 7.7, 5 Important

Ex 7.7, 6

Ex 7.7, 7 Important

Ex 7.7, 8 Important

Ex 7.7, 9

Ex 7.7, 10

Ex 7.7, 11 Important You are here

Ex 7.7, 12 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Ex 7.7, 13 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Ex 7.7, 14 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

Ex 7.7, 11 ∫1▒〖√(𝑥2 −8𝑥+7) " " 〗 𝑑𝑥 1/2 (x - 4)√(𝑥2 −8𝑥+7) + 9 log |𝑥−4+√(𝑥2−8𝑥+7)| + C 1/2 (x + 4)√(𝑥2 −8𝑥+7) + 9 log |𝑥+4+√(𝑥2−8𝑥+7)| + C 1/2 (x - 4)√(𝑥2 −8𝑥+7) - 3√2 log |𝑥−4+√(𝑥2−8𝑥+7)| + C 1/2 (x - 4)√(𝑥2 −8𝑥+7) + 9/2 log |𝑥−4+√(𝑥2−8𝑥+7)| + C ∫1▒〖√(𝑥^2−8𝑥+7) 𝑑𝑥〗 =∫1▒〖√(𝑥^2−2(4)(𝑥)+7) 𝑑𝑥〗 =∫1▒〖√(𝑥^2−2(4)(𝑥) 〖+(4)〗^2−(4)^2+7) 𝑑𝑥〗 =∫1▒〖√((𝑥−4)^2−16+7) 𝑑𝑥〗 =∫1▒〖√((𝑥−4)^2−9 ) 𝑑𝑥〗 =∫1▒〖√((𝑥−4)^2−(3)^2 ) 𝑑𝑥〗 =(𝑥 − 4)/2 √((𝑥−4)^2−(3)^2 )−(3)^2/2 𝑙𝑜𝑔|𝑥−4+√((𝑥−4)^2−(3)^2 )|+𝐶 =(𝑥 − 4)/2 √(𝑥^2−8𝑥+16−9 )−9/2 𝑙𝑜𝑔|𝑥−4+√(𝑥^2−8𝑥+16−9)|+𝐶 =(𝑥 − 4)/2 √(𝑥^2−8𝑥+7)−9/2 𝑙𝑜𝑔|𝑥−4+√(𝑥^2−8𝑥+7)|+𝐶 ∴ Option (D) is correct. It is of the form ∫1▒〖√(𝑥^2−𝑎^2 ) 𝑑𝑥=𝑥/2 √(𝑥^2−𝑎^2 )−𝑎^2/2 𝑙𝑜𝑔|𝑥+√(𝑥^2−𝑎^2 )|+𝐶〗 ∴ Replacing 𝑥 by 𝑥−4 and a by 3 , we get