# Ex 7.5, 6

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 7.5, 6 1 − 𝑥2𝑥(1 − 2𝑥) 1 − 𝑥2𝑥(1 − 2𝑥) = 1 − 𝑥2𝑥 − 2 𝑥2 Integrating 1 − 𝑥2𝑥 − 2 𝑥2 𝑑𝑥= 12 + − 𝑥2 + 1−2 𝑥2+ 𝑥 𝑑𝑥 = 12 𝑑𝑥+ 12 −𝑥 + 2−2 𝑥2+ 𝑥 𝑑𝑥 = 12 𝑑𝑥− 12 𝑥 − 2𝑥 1 − 2𝑥 𝑑𝑥 = 12 𝑑𝑥− 12 𝑥 − 2𝑥 2𝑥 − 1 𝑑𝑥 Now Solving 𝑥 − 2𝑥 2𝑥 − 1 = 𝐴𝑥 + 𝐵2𝑥 − 1 𝑥 − 2𝑥 2𝑥 − 1 = 𝐴 2𝑥 − 1 + 𝐵𝑥𝑥 2𝑥 − 1 Cancelling denominator 𝑥−2=𝐴 2𝑥−1+𝐵𝑥 Putting x = 0, in (2) 𝑥−2=𝐴 2𝑥−1+𝐵𝑥 0−2 = 𝐴 2×0−1 + 𝐵×0 −2 = A −1 −2 = −𝐴 𝐴 = 2 Similarly Putting x = 12 , in (2) 12 − 2 = A 12×2−1+𝐵× 12 − 32 = A 1−1+𝐵× 12 − 32 = A×0+ 𝐵2 − 32 = 𝐵2 𝐵 = −3 Hence we can write it as 𝑥 − 2𝑥 2𝑥 − 1 = 2𝑥 + −32𝑥 − 1 = 2𝑥 − 32𝑥 − 1 Therefore , from (1) we get, 1 − 𝑥2𝑥 1 − 2𝑥 = 12 𝑑𝑥+ 12 2𝑥 − 32𝑥 − 1 𝑑𝑥 = 12 𝑑𝑥+ 22 𝑑𝑥𝑥 − 32 𝑑𝑥2𝑥 − 1 = 12 𝑑𝑥+ 𝑑𝑥𝑥 + 32 𝑑𝑥1 − 2𝑥 = 12 𝑥+ log 𝑥+ 32 log 1 − 2𝑥−2 +𝐶 = 𝒙𝟐 + 𝒍𝒐𝒈 𝒙− 𝟑𝟒 𝒍𝒐𝒈 𝟏−𝟐𝒙+𝑪

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Ex 7.5, 6 You are here

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About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .