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Ex 7.5, 6 - Integrate 1 - x2 / x (1 - 2x) - Class 12 - Integration by partial fraction - Type 1

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.5, 6 1 − 𝑥2﷮𝑥(1 − 2𝑥)﷯ 1 − 𝑥2﷮𝑥(1 − 2𝑥)﷯ = 1 − 𝑥2﷮𝑥 − 2 𝑥﷮2﷯﷯ Integrating ﷮﷮ 1 − 𝑥﷮2﷯﷮𝑥 − 2 𝑥﷮2﷯﷯﷯ 𝑑𝑥= ﷮﷮ 1﷮2﷯ + − 𝑥﷮2﷯ + 1﷮−2 𝑥﷮2﷯+ 𝑥﷯﷯﷯ 𝑑𝑥 = 1﷮2﷯ ﷮﷮𝑑𝑥﷯+ 1﷮2﷯ ﷮﷮ −𝑥 + 2﷮−2 𝑥﷮2﷯+ 𝑥﷯﷯ 𝑑𝑥 = 1﷮2﷯ ﷮﷮𝑑𝑥﷯− 1﷮2﷯ ﷮﷮ 𝑥 − 2﷮𝑥 1 − 2𝑥﷯﷯﷯ 𝑑𝑥 = 1﷮2﷯ ﷮﷮𝑑𝑥﷯− 1﷮2﷯ ﷮﷮ 𝑥 − 2﷮𝑥 2𝑥 − 1﷯﷯﷯ 𝑑𝑥 Now Solving 𝑥 − 2﷮𝑥 2𝑥 − 1﷯﷯ = 𝐴﷮𝑥﷯ + 𝐵﷮2𝑥 − 1﷯ 𝑥 − 2﷮𝑥 2𝑥 − 1﷯﷯ = 𝐴 2𝑥 − 1﷯ + 𝐵𝑥﷮𝑥 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 = 1﷮2﷯ , in (2) 1﷮2﷯ − 2 = A 1﷮2﷯×2−1﷯+𝐵× 1﷮2﷯ − 3﷮2﷯ = A 1−1﷯+𝐵× 1﷮2﷯ − 3﷮2﷯ = A×0+ 𝐵﷮2﷯ − 3﷮2﷯ = 𝐵﷮2﷯ 𝐵 = −3 Hence we can write it as 𝑥 − 2﷮𝑥 2𝑥 − 1﷯﷯ = 2﷮𝑥﷯ + −3﷯﷮2𝑥 − 1﷯ = 2﷮𝑥﷯ − 3﷮2𝑥 − 1﷯ Therefore , from (1) we get, ﷮﷮ 1 − 𝑥﷮2﷯﷮𝑥 1 − 2𝑥﷯﷯﷯ = 1﷮2﷯ ﷮﷮𝑑𝑥﷯+ 1﷮2﷯ ﷮﷮ 2﷮𝑥﷯ − 3﷮2𝑥 − 1﷯﷯﷯ 𝑑𝑥 = 1﷮2﷯ ﷮﷮𝑑𝑥﷯+ 2﷮2﷯ ﷮﷮ 𝑑𝑥﷮𝑥﷯ − 3﷮2﷯﷯ ﷮﷮ 𝑑𝑥﷮2𝑥 − 1﷯﷯ = 1﷮2﷯ ﷮﷮𝑑𝑥﷯+ ﷮﷮ 𝑑𝑥﷮𝑥﷯ + 3﷮2﷯﷯ ﷮﷮ 𝑑𝑥﷮1 − 2𝑥﷯﷯ = 1﷮2﷯ 𝑥+ log﷮ 𝑥﷯﷯+ 3﷮2﷯ log﷮ 1 − 2𝑥﷯﷯﷮−2﷯ +𝐶 = 𝒙﷮𝟐﷯ + 𝒍𝒐𝒈﷮ 𝒙﷯﷯− 𝟑﷮𝟒﷯ 𝒍𝒐𝒈﷮ 𝟏−𝟐𝒙﷯﷯+𝑪

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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