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Ex 7.4, 17 - Integrate x + 2 / root x2 - 1 - Chapter 7 - Ex 7.4

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.4, 17 𝑥 + 2﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯ ﷮﷮ 𝑥 + 2﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ . 𝑑𝑥= ﷮﷮ 1﷮2﷯ 2𝑥﷯ + 2 ﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ . 𝑑𝑥 = ﷮﷮ 1﷮2﷯ 2𝑥﷯﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ . 𝑑𝑥+ ﷮﷮ 2﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ . 𝑑𝑥 = 1﷮2﷯ ﷮﷮ 2𝑥﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ . 𝑑𝑥+ ﷮﷮ 2﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ . 𝑑𝑥 Solving 𝑰𝟏 I1= 1﷮2﷯ ﷮﷮ 2𝑥﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ 𝑑𝑥 Let 𝑥﷮2﷯−1=𝑡 Differentiating w.r.t. x 2𝑥−0= 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥= 𝑑𝑡﷮2𝑥﷯ Thus, our equation becomes ∴ I1= 1﷮2﷯ ﷮﷮ 2𝑥﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ 𝑑𝑥 Put the values of 𝑥﷮2﷯−1﷯=𝑡 and 𝑑𝑥, we get I1= 1﷮2﷯ ﷮﷮ 2𝑥﷮ ﷮𝑡﷯﷯﷯ 𝑑𝑥 I1= 1﷮2﷯ ﷮﷮ 2𝑥﷮ ﷮𝑡﷯﷯﷯ × 𝑑𝑡﷮2𝑥﷯ I1= 1﷮2﷯ ﷮﷮ 1﷮ ﷮𝑡﷯﷯﷯ 𝑑𝑡 I1= 1﷮2﷯ ﷮﷮ 1﷮ 𝑡﷮ 1﷮2﷯﷯﷯﷯ 𝑑𝑡 I1= 1﷮2﷯ ﷮﷮ 𝑡﷮ −1﷮2﷯﷯﷯ 𝑑𝑡 I1= 1﷮2﷯ 𝑡﷮ −1﷮2﷯ + 1﷯﷮ −1﷮2﷯ + 1﷯ +𝐶1 I1= 1﷮2﷯ 𝑡﷮ 1﷮2﷯﷯﷮ 1﷮2﷯﷯ +𝐶1 I1= 𝑡﷮ 1﷮2﷯﷯+𝐶1 I1= ﷮𝑡﷯+𝐶1 I1= ﷮ 𝑥﷮2﷯ − 1﷯ + 𝐶1 Solving 𝑰𝟐 I2= ﷮﷮ 2﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ . 𝑑𝑥 I2=2 ﷮﷮ 1﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷮2﷯﷯﷯﷯ . 𝑑𝑥 I2=2 𝑙𝑜𝑔﷮ 𝑥+ ﷮ 𝑥﷮2﷯ − 1﷯﷮2﷯﷯﷯﷯+𝐶 Now, putting the values of I1 and I2 in (1) ﷮﷮ 𝑥 + 2﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ . 𝑑𝑥= 1﷮2﷯ ﷮﷮ 2𝑥﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ . 𝑑𝑥+2 ﷮﷮ 1﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ . 𝑑𝑥 = ﷮ 𝑥﷮2﷯ − 1﷯ + 𝐶1+2 𝑙𝑜𝑔﷮ 𝑥+ ﷮ 𝑥﷮2﷯ −1﷯ ﷯﷯+𝐶3 = ﷮ 𝒙﷮𝟐﷯ − 𝟏﷯+𝟐 𝒍𝒐𝒈﷮ 𝒙+ ﷮ 𝒙﷮𝟐﷯ −𝟏﷯﷯﷯+ 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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