# Ex 7.5, 21 - Chapter 7 Class 12 Integrals

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 7.5, 21 1( 𝑒𝑥 − 1) [Hint : Put ex = t] Let 𝑒𝑥 = 𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑒𝑥 = 𝑑𝑡𝑑𝑥 𝑑𝑥 = 𝑑𝑡 𝑒𝑥 Therefore 1( 𝑒𝑥 − 1) 𝑑𝑥 = 1 𝑡 − 1 𝑑𝑡 𝑒𝑥 We can write integrand as 1𝑡 𝑡 − 1 = 𝐴𝑡 + 𝐵𝑡 − 1 1𝑡 𝑡 − 1 = 𝐴 𝑡 − 1 + 𝐵 𝑡𝑡 𝑡 − 1 By cancelling denominator 1 = 𝐴 𝑡−1+𝐵 𝑡 Putting t = 0 in (1) 1 = 𝐴 0−1+𝐵×0 1 = 𝐴× −1 1 = −𝐴 𝐴 = −1 Similarly putting t = 1 1 = 𝐴 1−1+𝐵×1 1 = 𝐴×0+𝐵 1 = 𝐵 𝐵 = 1 Therefore 1𝑡 𝑡 − 1 𝑑𝑡 = −1𝑡 𝑑𝑡 + 1𝑡 − 1 = − log 𝑡+ log 𝑡−1+𝐶 = log 𝑡 − 1𝑡+𝐶 Putting t = 𝑒𝑥 1 𝑒𝑥 − 1 𝑑𝑥 = 𝑙𝑜𝑔 𝑒𝑥 − 1 𝑒𝑥+𝐶 = 𝐥𝐨𝐠 𝒆𝒙 − 𝟏 𝒆𝒙+𝑪

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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.