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Ex 7.2, 19 - Integrate e2x - 1 / e2x + 1 - Chapter 7 - Ex 7.2

Ex 7.2, 19 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.2, 19 - Chapter 7 Class 12 Integrals - Part 3

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Transcript

Ex 7.2, 19 Integrate the function (𝑒2𝑥 − 1)/(𝑒2𝑥+ 1) Simplify the given function (𝑒^2𝑥 − 1)/(𝑒^2𝑥 + 1) Dividing numerator and denominator by ex, we obtain = (𝑒^2𝑥/𝑒^𝑥 " " −" " 𝟏/𝒆^𝒙 )/(𝑒^2𝑥/𝑒^𝑥 " " + " " 𝟏/𝒆^𝒙 ) = (𝑒^𝒙 − 𝒆^(−𝒙))/(𝑒^𝒙 + 𝒆^(−𝒙) ) Let 𝑒^𝑥 + 𝑒^(−𝑥)= 𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑒^𝑥+(−1) 𝑒^(−𝑥)= 𝑑𝑡/𝑑𝑥 𝑒^𝑥−𝑒^(−𝑥)= 𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(𝑒^𝑥 − 𝑒^(−𝑥) ) Now, Integrating the function ∫1▒〖" " (𝑒^2𝑥 − 1)/(𝑒^2𝑥 + 1) " " 〗. 𝑑𝑥 = ∫1▒〖" " (𝑒^𝑥 − 𝑒^(−𝑥))/(𝑒^𝑥 + 𝑒^(−𝑥) ) " " 〗. 𝑑𝑥 (Using (𝑒^2𝑥 − 1)/(𝑒^2𝑥 + 1)=(𝑒^𝑥 − 𝑒^(−𝑥))/(𝑒^𝑥 + 𝑒^(−𝑥) ) ) Putting 𝑒^𝑥 + 𝑒^(−𝑥)=𝑡 & 𝑑𝑥=𝑑𝑡/(𝑒^𝑥 − 𝑒^(−𝑥) ) =∫1▒〖" " (𝑒^𝑥 − 𝑒^(−𝑥))/𝑡 " " 〗. 𝑑𝑡/(𝑒^𝑥 − 𝑒^(−𝑥) ) " " =∫1▒〖" " 1/𝑡 " " 〗. 𝑑𝑡 =log⁡|𝑡|+𝐶 =log⁡〖 |𝑒^𝑥+𝑒^(−𝑥) |〗+𝐶 =𝒍𝒐𝒈⁡(𝒆^𝒙 + 𝒆^(−𝒙) )+𝑪 (Using 𝑡=𝑒^𝑥 + 𝑒^(−𝑥)) (As 𝑒^𝑥+𝑒^(−𝑥)>0 )

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Davneet Singh

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.