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Ex 7.4

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Ex 7.4, 7 You are here

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Ex 7.4, 24 (MCQ)

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Last updated at May 29, 2023 by Teachoo

Ex 7.4, 7 𝑥 − 1 𝑥2 − 1 Integrating the function 𝑤.𝑟.𝑡.𝑥 𝑥 − 1 𝑥2 − 1 𝑑𝑥 = 𝑥 𝑥2 − 1 − 1 𝑥2 − 1 𝑑𝑥 = 𝑥 𝑥2 − 1 𝑑𝑥− 1 𝑥2 − 1 𝑑𝑥 Solving 𝐈𝟏 I1 = 𝑥 𝑥2 − 1 𝑑𝑥 Let 𝑥2 − 1=𝑡 Diff both sides w.r.t.x 2𝑥−0= 𝑑𝑡𝑑𝑥 𝑑𝑥= 𝑑𝑡2𝑥 Thus, our equation becomes ∴ I1 = 𝑥 𝑥2 − 1 𝑑𝑥 Put the values of 𝑥2 −1=𝑡 and 𝑑𝑥= 𝑑𝑡2𝑥 I1 = 𝑥 𝑡 . 𝑑𝑡2𝑥 I1 = 12 1 𝑡 𝑑𝑡 I1 = 12 1 𝑡 12 𝑑𝑡 I1 = 12 𝑡 −12 𝑑𝑡 I1 = 12 𝑡 −12 + 1 −12 + 1+𝐶1 I1 = 12 . 𝑡 12 12+𝐶1 I1 = 𝑡 12+𝐶1 I1 = 𝑡+𝐶1 I1 = 𝑥2−1 +𝐶1 Solving 𝐈𝟐 I2 = 1 𝑥2 − 1 𝑑𝑥 I2 = 1 𝑥2 − 12 𝑑𝑥 I2 = log 𝑥+ 𝑥2 − 12+𝐶2 I2 = log 𝑥+ 𝑥2 −1 +𝐶2 Now, putting the value of I1 and I2 in eq. (1) 𝑥 − 1 𝑥2 − 1 𝑑𝑥= 𝑥 𝑥2 − 1 𝑑𝑥− 1 𝑥2 − 1 𝑑𝑥 = 𝑥2−1 +𝐶1− log 𝑥+ 𝑥2 −1 +𝐶2 = 𝑥2−1 +𝐶1− log 𝑥+ 𝑥2 −1 −𝐶2 = 𝒙𝟐−𝟏 − 𝒍𝒐𝒈 𝒙+ 𝒙𝟐 −𝟏 −𝑪