Misc 38 (MCQ) - Chapter 7 Class 12 Integrals

Last updated at April 16, 2024 by

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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).

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