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Ex 7.5 - i.jpg

Ex 7.5 ii.jpg
Ex 7.5 iii.jpg Ex 7.5 iV.jpg

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Transcript

Ex 7.5, 18 𝑥2+ 1﷯ 𝑥2+ 2﷯﷮ 𝑥2+ 3﷯ 𝑥2+ 4﷯﷯ Let t = 𝑥﷮2﷯ 𝑥2 + 1﷯ 𝑥2 + 2﷯﷮ 𝑥2 + 3﷯ 𝑥2 + 4﷯﷯ = 𝑡 + 1﷯ 𝑡 + 2﷯﷮ 𝑡 + 3﷯ 𝑡 + 4﷯﷯ = 𝑡﷮2﷯ + 3𝑡 + 2﷮ 𝑡﷮2﷯ + 7𝑡 + 12﷯ = 1 + −4𝑡 −10﷮ 𝑡﷮2﷯ + 7𝑡 + 12﷯ = 1 + −4𝑡 −10﷮ 𝑡 + 3﷯ 𝑡 + 4﷯﷯ We can write 4𝑡 + 10﷮ 𝑡 + 3﷯ 𝑡 + 4﷯﷯ = 𝐴﷮ 𝑡 + 3﷯﷯ + 𝐵﷮ 𝑡 + 4﷯﷯ 4𝑡 + 10﷮ 𝑡 + 3﷯ 𝑡 + 4﷯﷯ = 𝐴 𝑡 + 4﷯ + 𝐵 𝑡 + 3﷯﷮ 𝑡 + 3﷯ 𝑡 + 4﷯﷯ By cancelling denominator 4𝑡−10 = 𝐴 𝑡+4﷯+𝐵 𝑡+3﷯ Putting t = − 4 in (1) 4 −4﷯+10 = 𝐴 −4+4﷯+𝐵 −4+3﷯ −16+10 = 𝐴×0+𝐵 −1﷯ −6 = 𝐴×0+𝐵 −1﷯ −6 = −𝐵 𝐵 = 6 Similarly putting t = −3 in (1) 4 −3﷯+10 = 𝐴 −3+4﷯+𝐵 −3+3﷯ −12+10 = 𝐴×1+𝐵×0 −2 = 𝐴 𝐴 = −2 Hence we can write 4𝑡 + 10﷮ 𝑡 + 3﷯ 𝑡 + 4﷯﷯ = −2﷮ 𝑡 + 3﷯﷯ + 6﷮ 𝑡 + 4﷯﷯ Putting back t = 𝑥﷮2﷯ 4 𝑥﷮2﷯ − 10﷮ 𝑥﷮2﷯ + 3﷯ 𝑥﷮2﷯ + 4﷯﷯ = −2﷮ 𝑥﷮2﷯ + 3﷯﷯ + 6﷮ 𝑥﷮2﷯ + 4﷯﷯ Therefore ﷮﷮ 𝑥2+ 1﷯ 𝑥2+ 2﷯﷮ 𝑥2+ 3﷯ 𝑥2+ 4﷯﷯﷯ = ﷮﷮1− −2﷮ 𝑥﷮2﷯ + 3﷯﷯ + 6﷮ 𝑥﷮2﷯ + 4﷯﷯﷯﷯ 𝑑𝑥 = ﷮﷮1﷯. 𝑑𝑥 + ﷮﷮ 2﷮ 𝑥﷮2﷯ + 3﷯﷯﷯ 𝑑𝑥 − ﷮﷮ 6﷮ 𝑥﷮2﷯ + 4﷯﷯﷯ 𝑑𝑥 = ﷮﷮1﷯. 𝑑𝑥 + 2 ﷮﷮ 1﷮ 𝑥﷮2﷯ + ﷮3﷯﷯﷮2﷯﷯﷯ 𝑑𝑥 − 6 ﷮﷮ 1﷮ 𝑥﷮2﷯ + 2﷮2﷯﷯﷯﷯ 𝑑𝑥 = 𝑥 + 2 × 1﷮ ﷮3﷯﷯ tan﷮−1﷯﷮ 𝑥﷮ ﷮3﷯﷯﷯ − 6 × 1﷮2﷯ tan﷮−1﷯﷮ 𝑥﷮2﷯﷯+𝐶 = 𝒙 + 𝟐﷮ ﷮𝟑﷯﷯ 𝒕𝒂𝒏﷮−𝟏﷯﷮ 𝒙﷮ ﷮𝟑﷯﷯﷯﷯−𝟑 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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