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

Share on WhatsApp

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)

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo