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Misc 14 - Integrate 1 / (x2 + 1) (x2 + 4) - Chapter 7 - Miscellaneous

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Misc 14 Integrate the function 1﷮ 𝑥﷮2﷯ + 1 ﷯ 𝑥﷮2﷯ + 4 ﷯ ﷯ Integrate 1﷮ 𝑥﷮2﷯ + 1 ﷯ 𝑥﷮2﷯ + 4 ﷯ ﷯ Let 𝑥﷮2﷯=𝑦 1﷮ 𝑥﷮2﷯ + 1 ﷯ 𝑥﷮2﷯ + 4 ﷯ ﷯= 1﷮ 𝑦 + 4﷯ 𝑦 + 1﷯﷯ Hence we can write 1﷮ 𝑦 + 4﷯ 𝑦 + 1﷯﷯= 𝐴﷮ 𝑦 + 4﷯﷯+ 𝐵﷮ 𝑦 + 1﷯﷯ 1﷮ 𝑦 + 4﷯ 𝑦 + 1﷯﷯= 𝐴 𝑦 + 1﷯+𝐵 𝑦 + 4﷯﷮ 𝑦 + 4﷯ 𝑦 + 1﷯﷯ By canceling denominators 1 = A 𝑦+1﷯+B 𝑦+4﷯ Putting 𝑦=−1 1= A −1+1﷯+B −1+4﷯ 1= A ×0+ B 3﷯ 1=3B B = 1﷮3﷯ Similarly putting 𝑦=−4 1= A −4+1﷯+ B −4+4﷯ 1= A −3﷯+ B 0﷯ 1=−3A A = −1﷮3﷯ Therefore we can write 1﷮ 𝑦 + 4﷯ 𝑦 + 1﷯﷯= −1﷮ 3﷯﷯﷮ 𝑦 + 4﷯﷯+ 1﷮3﷯﷯﷮ 𝑦 + 1﷯﷯ Putting 𝑦= 𝑥﷮2﷯ 1﷮ 𝑥﷮2﷯ + 4﷯ 𝑥﷮2﷯ + 1﷯﷯= −1﷮ 3﷯﷯﷮ 𝑥﷮2﷯ + 4﷯﷯+ 1﷮ 3﷯﷯﷮ 𝑥﷮2﷯+ 1﷯﷯ = −1﷮3 𝑥﷮2﷯ + 4﷯﷯+ 1﷮3 𝑥﷮2 ﷯+ 1﷯﷯ Integrating w.r.t.𝑥 ﷮﷮ 1﷮ 𝑥﷮2 ﷯+ 4﷯ 𝑥﷮2﷯ + 1﷯﷯𝑑𝑥= ﷮﷮ −1﷮3 𝑥﷮2﷯ + 4﷯﷯+ 1﷮3 𝑥﷮2﷯ + 1﷯﷯﷯𝑑𝑥﷯﷯ = ﷮﷮ −1﷮3 𝑥﷮2﷯ + 4﷯﷯𝑑𝑥+﷯ ﷮﷮ −1﷮3 𝑥﷮2﷯ + 1﷯﷯𝑑𝑥﷯ = −1﷮ 3﷯ ﷮﷮ 1﷮ 𝑥﷮2﷯ + 2﷮2﷯﷯𝑑𝑥+ 1﷮3﷯ ﷮﷮ 1﷮ 𝑥﷮2﷯ + 1﷮2﷯﷯𝑑𝑥﷯﷯ = −1﷮3﷯ × 1﷮2﷯ tan﷮−1﷯﷮ 𝑥﷮2﷯+ 1﷮3﷯ × 1﷮1﷯ tan﷮−1﷯﷮ 𝑥﷮1﷯+𝐶﷯﷯ = −1﷮ 6﷯ tan﷮−1﷯﷮ 𝑥﷮2﷯+ 1﷮3﷯ tan﷮−1﷯﷮𝑥+𝑐﷯﷯ = 𝟏﷮𝟑﷯ 𝐭𝐚𝐧﷮−𝟏﷯﷮𝒙− 𝟏﷮𝟔﷯ 𝐭𝐚𝐧﷮−𝟏﷯﷮ 𝒙﷮𝟐﷯+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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