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Ex 7.2, 20 - Integrate (e2x - e-2x) / (e2x + e-2x) - Ex 7.2

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Ex 7.2, 20 (𝑒^2π‘₯ βˆ’ 𝑒^(βˆ’2π‘₯))/(𝑒^2π‘₯ + 𝑒^(βˆ’2π‘₯) ) Step 1: Let 𝑒^2π‘₯ + 𝑒^(βˆ’2π‘₯)= 𝑑 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑒^2π‘₯. 𝑑(2π‘₯)/𝑑π‘₯ +𝑒^(βˆ’2π‘₯) 𝑑(βˆ’2π‘₯)/𝑑π‘₯= 𝑑𝑑/𝑑π‘₯ γ€–2𝑒〗^2π‘₯βˆ’γ€–2𝑒〗^(βˆ’2π‘₯)= 𝑑𝑑/𝑑π‘₯ 2(𝑒^2π‘₯βˆ’π‘’^(βˆ’2π‘₯) )=𝑑𝑑/𝑑π‘₯ " " 𝑑π‘₯ = 𝑑𝑑/2(𝑒^2π‘₯βˆ’ 𝑒^(βˆ’2π‘₯) ) Step 2: Integrating the function ∫1β–’γ€–" " (𝑒^2π‘₯ βˆ’ 𝑒^(βˆ’2π‘₯))/(𝑒^2π‘₯ + 𝑒^(βˆ’2π‘₯) )γ€—. 𝑑π‘₯ Putting 𝑒^2π‘₯ + 𝑒^(βˆ’2π‘₯)=𝑑 & 𝑑π‘₯=𝑑𝑑/2(𝑒^2π‘₯βˆ’ 𝑒^(βˆ’2π‘₯) ) = ∫1β–’γ€–" " (𝑒^2π‘₯ βˆ’ 𝑒^(βˆ’2π‘₯))/𝑑〗. 𝑑𝑑/2(𝑒^2π‘₯βˆ’ 𝑒^(βˆ’2π‘₯) ) = ∫1β–’γ€–" " 1/2𝑑〗. 𝑑𝑑 = 1/2 ∫1β–’1/𝑑. 𝑑𝑑 = 1/2 log⁑〖 |𝑑|γ€—+𝐢 = 1/2 log⁑〖 |𝑒^2π‘₯ + 𝑒^(βˆ’2π‘₯) |γ€—+𝐢 = 𝟏/𝟐 π’π’π’ˆβ‘γ€– (𝒆^πŸπ’™ + 𝒆^(βˆ’πŸπ’™) )γ€—+π‘ͺ

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