Ex 7.2, 19 - Chapter 7 Class 12 Integrals
Last updated at Dec. 6, 2019 by Teachoo
Transcript
Ex 7.2, 19 𝑒2𝑥 −1𝑒2𝑥+1 Step 1: Simplify the given function 𝑒2𝑥 − 1 𝑒2𝑥 + 1 Dividing numerator and denominator by ex, we obtain = 𝑒2𝑥 𝑒𝑥 − 𝟏 𝒆𝒙 𝑒2𝑥 𝑒𝑥 + 𝟏 𝒆𝒙2 = 𝑒𝟐𝒙 −𝒙 − 1 . 𝒆−𝒙 𝑒𝟐𝒙 −𝒙 + 1 . 𝒆−𝒙 = 𝑒𝒙 − 𝒆−𝒙 𝑒𝒙 + 𝒆−𝒙 ∴ 𝑒2𝑥 − 1 𝑒2𝑥 + 1 = 𝑒𝑥 − 𝑒−𝑥 𝑒𝑥 + 𝑒−𝑥 Step 2: Let 𝑒𝑥 + 𝑒−𝑥= 𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑒𝑥+ −1 𝑒−𝑥= 𝑑𝑡𝑑𝑥 𝑒𝑥− 𝑒−𝑥= 𝑑𝑡𝑑𝑥 𝑑𝑥= 𝑑𝑡 𝑒𝑥 − 𝑒−𝑥 Step 3: Integrating the function 𝑒2𝑥 − 1 𝑒2𝑥 + 1 . 𝑑𝑥 = 𝑒𝑥 − 𝑒−𝑥 𝑒𝑥 + 𝑒−𝑥 . 𝑑𝑥 Putting 𝑒𝑥 + 𝑒−𝑥=𝑡 & 𝑑𝑥= 𝑑𝑡 𝑒𝑥 − 𝑒−𝑥 = 𝑒𝑥 − 𝑒−𝑥𝑡 . 𝑑𝑡 𝑒𝑥 − 𝑒−𝑥 = 1𝑡 . 𝑑𝑡 = log 𝑡+𝐶 = log 𝑒𝑥+ 𝑒−𝑥+𝐶 = 𝒍𝒐𝒈 𝒆𝒙 + 𝒆−𝒙+𝑪
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