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Misc 38 βˆ«β–’π‘‘π‘₯/(𝑒^π‘₯ + 𝑒^(βˆ’π‘₯) ) is equal to (A) tan^(βˆ’1) (𝑒^π‘₯ )+𝐢 (B) tan^(βˆ’1)⁑〖(𝑒^(βˆ’π‘₯) )+𝐢〗 (C) log⁑(𝑒^π‘₯βˆ’π‘’^(βˆ’π‘₯) )+𝐢 (D) log⁑(𝑒^π‘₯+𝑒^(βˆ’π‘₯) )+𝐢 βˆ«β–’π‘‘π‘₯/(𝑒^π‘₯ + 𝑒^(βˆ’π‘₯) ) = βˆ«β–’π‘‘π‘₯/(𝑒^π‘₯ + 1/𝑒^π‘₯ ) = ∫1β–’(𝑒^π‘₯ 𝑑π‘₯)/(𝑒^2π‘₯ + 1) Let 𝑒^π‘₯=𝑑 𝑑𝑑/𝑑π‘₯=𝑒^π‘₯ dt = 𝑒^π‘₯ 𝑑π‘₯ Substituting, = ∫1▒𝑑𝑑/(𝑑^2 +1) = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (𝑑)+ C Putting value of t = 〖𝒕𝒂𝒏〗^(βˆ’πŸ) (𝒆^𝒙 )+ C Hence, answer is (A).

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo