Ex 7.5, 21 - Integrate 1 / (ex - 1) [Hing: Put ex = t] - Ex 7.5

Ex 7.5, 21 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.5, 21 - Chapter 7 Class 12 Integrals - Part 3

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Transcript

Ex 7.5, 21 Integrate the function 1/((𝑒^𝑥 − 1) ) [Hint : Put ex = t] Let 𝑒^𝑥 = 𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑒^𝑥 = 𝑑𝑡/𝑑𝑥 𝑑𝑥 = 𝑑𝑡/𝑒^𝑥 Therefore ∫1▒1/((𝑒^𝑥 − 1) ) 𝑑𝑥 = ∫1▒1/((𝑡 − 1) ) 𝑑𝑡/𝑒^𝑥 = ∫1▒𝑑𝑡/(𝑡(𝑡 − 1) ) We can write integrand as 1/(𝑡(𝑡 − 1) ) = 𝐴/𝑡 + 𝐵/(𝑡 − 1) 1/(𝑡(𝑡 − 1) ) = (𝐴(𝑡 − 1) + 𝐵𝑡)/𝑡(𝑡 − 1) Cancelling denominator 1 = 𝐴(𝑡−1)+𝐵𝑡 Putting t = 0 in (1) 1 = 𝐴(0−1)+𝐵×0 1 = 𝐴×(−1) 1 = −𝐴 𝐴 = −1 Putting t = 1 1 = 𝐴(1−1)+𝐵×1 1 = 𝐴×0+𝐵 1 = 𝐵 𝐵 = 1 Therefore ∫1▒1/(𝑡(𝑡 − 1) ) 𝑑𝑡 = ∫1▒(−1)/(𝑡 ) 𝑑𝑡 + ∫1▒1/(𝑡 − 1 ) = −〖log 〗⁡|𝑡|+〖log 〗⁡|𝑡−1|+𝐶 = 〖log 〗⁡|(𝑡 − 1)/𝑡|+𝐶 Putting back t = 𝑒^𝑥 = 〖𝑙𝑜𝑔 〗⁡|(𝑒^𝑥 − 1)/𝑒^𝑥 |+𝐶 = 〖𝐥𝐨𝐠 〗⁡((𝒆^𝒙 − 𝟏)/𝒆^𝒙 )+𝑪 Since ex > 1 for x > 0 ∴ ex – 1 > 0 ⇒ |(𝒆^𝒙 − 𝟏)/𝒆^𝒙 |=((𝒆^𝒙 − 𝟏)/𝒆^𝒙 ) ("As " 𝑙𝑜𝑔 𝐴−𝑙𝑜𝑔 𝐵" = " 𝑙𝑜𝑔 𝐴/𝐵)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.