# Ex 7.5, 23 - Chapter 7 Class 12 Integrals

Last updated at July 11, 2018 by Teachoo

Last updated at July 11, 2018 by Teachoo

Transcript

Ex 7.5, 23 𝑑𝑥𝑥(𝑥2 + 1) equals • log 𝑥 - 12 log (x2+1) + C • log 𝑥 + 12 log (x2+1) + C • log 𝑥 + 12 log (x2+1) + C • 12 log 𝑥 + log (x2+1) + C By Multiplying integrand by 𝑥𝑥 1𝑥(𝑥2 + 1) . 𝑥𝑥 𝑑𝑥 = 𝑥 𝑑𝑥 𝑥2 𝑥2 + 1 Let t = 𝑥2 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 2𝑥 = 𝑑𝑡𝑑𝑥 𝑑𝑥 = 𝑑𝑡2𝑥 Thus, our equation become 𝑥 𝑥2( 𝑥2 + 1) 𝑑𝑥 = 𝑥𝑡(𝑡 + 1) 𝑑𝑡2𝑥 = 12 𝑑𝑡𝑡(𝑡 + 1) = 12 𝑑𝑡𝑡(𝑡 + 1) We can write integrand as 1𝑡(𝑡 + 1) = 𝐴𝑡 + 𝐵𝑡 + 1 1𝑡(𝑡 + 1) = 𝐴(𝑡 + 1) + 𝐵𝑡𝑡(𝑡 + 1) By cancelling denominator 1 = 𝐴(𝑡+1)+𝐵𝑡 Putting t = 0 in (1) 1 = 𝐴(0+1)+𝐵×0 1 = 𝐴 𝐴 = 1 Similarly putting x = −1 in (1) 1 = 𝐴(−1+1)+𝐵 −1 1 = 𝐴×0−𝐵 1 = −𝐵 𝐵 = −1 Therefore 12 1𝑡 𝑡 + 1 𝑑𝑡 = 12 1𝑡 𝑑𝑡 + 12 −1𝑡 + 1 𝑑𝑡 = 12 log 𝑡− 12 log 𝑡+1+𝐶 Putting back t = 𝑥2 = 12 log 𝑥2− 12 log 𝑥2+1+𝐶 = 12×2 log 𝑥− 12 log 𝑥2+1+𝐶 = log 𝑥− 12 log ( 𝑥2+1)+𝐶 ∴ The correct answer is A .

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Ex 7.5, 23 You are here

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.