Misc 41 ∫▒𝑑𝑥/(𝑒^𝑥 + 𝑒^(−𝑥) ) is equal to (A) tan^(−1) (𝑒^𝑥 )+𝐶 (B) tan^(−1)⁡〖(𝑒^(−𝑥) )+𝐶〗 (C) log⁡(𝑒^𝑥−𝑒^(−𝑥) )+𝐶 (D) log⁡(𝑒^𝑥+𝑒^(−𝑥) )+𝐶 ∫▒𝑑𝑥/(𝑒^𝑥 + 𝑒^(−𝑥) ) = ∫▒𝑑𝑥/(𝑒^𝑥 + 1/𝑒^𝑥 ) = ∫1▒(𝑒^𝑥 𝑑𝑥)/(𝑒^2𝑥 + 1) Let 𝑒^𝑥=𝑡 𝑑𝑡/𝑑𝑥=𝑒^𝑥 dt = 𝑒^𝑥 𝑑𝑥 Substituting, = ∫1▒𝑑𝑡/(𝑡^2 +1) = 〖𝑡𝑎𝑛〗^(−1) (𝑡)+ C Putting value of t = 〖𝒕𝒂𝒏〗^(−𝟏) (𝒆^𝒙 )+ C Hence, answer is (A). (∫1▒〖𝑑𝑥/(𝑥^2 + 1)=〖𝑡𝑎𝑛〗^(−1) 𝑥〗 " " )

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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