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Ex 7.5, 13 - Integrate 2 / (1 - x) (1 + x2) - Chapter 7 NCERT - Integration by partial fraction - Type 5

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.5, 13 2﷮ 1 − 𝑥﷯ 1 + 𝑥﷮2﷯﷯﷯ We can write the integrand as 2﷮ 1 − 𝑥﷯ 1 + 𝑥﷮2﷯﷯﷯ = 2﷮− 𝑥 − 1﷯ 1 + 𝑥﷮2﷯﷯﷯ = −2﷮ 𝑥 − 1﷯ 1 + 𝑥﷮2﷯﷯﷯ Let −2﷮ 𝑥 − 1﷯ 1 + 𝑥﷮2﷯﷯﷯ = 𝐴﷮ 𝑥 − 1﷯﷯ + 𝐵𝑥 + 𝐶﷮ 1 + 𝑥﷮2﷯﷯﷯ −2﷮ 𝑥 − 1﷯ 1 + 𝑥﷮2﷯﷯﷯ = 𝐴 1 + 𝑥﷮2﷯﷯ + 𝐵𝑥 + 𝐶﷯ 𝑥 − 1﷯﷮ 𝑥 − 1﷯ 1 + 𝑥﷮2﷯﷯﷯ By cancelling denominator −2 = 𝐴 1+ 𝑥﷮2﷯﷯ + 𝐵𝑥+𝐶﷯ 𝑥−1﷯ Putting x = 1 , in (1) −2 = A 1+ 1﷮2﷯﷯ + Bx + C﷯ 1−1﷯ −2 = A 1+1﷯+ Bx+C﷯ 0 −2 = 2A+0 −2﷮2﷯ = A −1 = A ∴ A = −1 Putting x = 0 , in (1) −2 = A 1+ 0﷮2﷯﷯ + B 0﷯+C﷯ 0−1﷯ −2 = A 1﷯ +C −1﷯ −2 = A−C Putting A = −1 −2 = −1−C C = −1+2 C = 1 Now, Putting A = −1 , in (1) −2 = A 1+1﷯ + B −1﷯+C﷯ −1−1﷯ −2 = 2𝐴+ −𝐵+𝐶﷯ −2﷯ −2 = 2𝐴+2𝐵−2𝐶 −2 = −2+2𝐵−2 −2+4=2𝐵 2𝐵=2 𝐵=1 Thus, 𝐴=−1, 𝐵=1, 𝐶 = 1 So, we can write −2﷮ 𝑥 − 1﷯ 1 + 𝑥﷮2﷯﷯﷯ = 𝐴﷮ 𝑥 − 1﷯﷯ + 𝐵𝑥 + 𝐶﷮ 1 + 𝑥﷮2﷯﷯﷯ = −1﷮ 𝑥 − 1﷯﷯ + 𝑥 + 1﷮ 𝑥﷮2﷯ + 1﷯﷯ Therefore integrating ﷮﷮ −2﷮ 1 − 𝑥﷯ 1 + 𝑥﷮2﷯﷯﷯﷯ 𝑑𝑥 = ﷮﷮ −1﷮𝑥 − 1﷯﷯ 𝑑𝑥 + ﷮﷮ 𝑥 + 1﷮ 𝑥﷮2﷯ + 1﷯﷯ 𝑑𝑥 = − ﷮﷮ 1﷮𝑥 − 1﷯﷯ 𝑑𝑥 + ﷮﷮ 𝑥﷮ 𝑥﷮2﷯ + 1﷯﷯ 𝑑𝑥 + ﷮﷮ 1﷮ 𝑥﷮2﷯ + 1﷯﷯ 𝑑𝑥 Solving 𝐈1 I2 = ﷮﷮ 𝑥﷮ 𝑥﷮2﷯ + 1﷯﷯ 𝑑𝑥 Let 𝑡 = 𝑥﷮2﷯+1 𝑑𝑡﷮𝑑𝑥﷯ = 2𝑥 𝑑𝑡﷮2𝑥﷯ = 𝑑𝑥 Hence ﷮﷮ 𝑥﷮ 𝑥﷮2﷯ + 1﷯﷯ 𝑑𝑥 = ﷮﷮ 𝑥﷮𝑡﷯ . 𝑑𝑡﷮2𝑥﷯﷯ = ﷮﷮ 𝑑𝑡﷮2 𝑡﷯﷯﷯ = 1﷮2﷯ log ﷮ 𝑡﷯﷯+𝐶1 Putting back t = 𝑥﷮2﷯+1 = 1﷮2﷯ log ﷮ 𝑥﷮2﷯+1﷯﷯+𝐶2 Therefore ﷮﷮ 2﷮ 1 − 𝑥﷯ 1 + 𝑥﷮2﷯﷯﷯﷯ 𝑑𝑥 = ﷮﷮ −1﷮𝑥 − 1﷯﷯ 𝑑𝑥 + ﷮﷮ 𝑥﷮ 𝑥﷮2﷯ + 1﷯﷯ 𝑑𝑥 + ﷮﷮ 1﷮ 𝑥﷮2﷯ + 1﷯﷯ 𝑑𝑥 = − log ﷮ 𝑥−1﷯﷯+ 1﷮2﷯ log ﷮ 𝑥﷮2﷯+1﷯﷯+ tan﷮−1﷯﷮𝑥﷯+𝐶 = − 𝒍𝒐𝒈 ﷮ 𝒙−𝟏﷯﷯+ 𝟏﷮𝟐﷯ 𝒍𝒐𝒈 ﷮ 𝒙﷮𝟐﷯+𝟏﷯﷯+ 𝒕𝒂𝒏﷮−𝟏﷯﷮𝒙﷯+𝑪

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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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