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Ex 7.4, 22 - Integrate x + 3 / x2 + 2x - 5 - Ex 7.4

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.4, 22 Solving 𝑰𝟏 I1= 1﷮2﷯ ﷮﷮ 2𝑥 − 2﷮ 𝑥﷮2﷯ − 2𝑥 − 5﷯﷯ . 𝑑𝑥 Let 𝑥﷮2﷯ − 2𝑥 − 5=𝑡 Diff both sides w.r.t.x 2𝑥−2−0= 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥= 𝑑𝑡﷮2𝑥 − 2﷯ Thus, our equation becomes ∴ I1= 1﷮2﷯ ﷮﷮ 2𝑥 − 2﷮ 𝑥﷮2﷯ − 2𝑥 − 5﷯﷯ . 𝑑𝑥 Putting value of 𝑥﷮2﷯−2𝑥−5﷯=𝑡 and 𝑑𝑥= 𝑑𝑡﷮2𝑥 − 2﷯ I1= 1﷮2﷯ ﷮﷮ 2𝑥 − 2﷮𝑡﷯﷯ . 𝑑𝑥 I1= 1﷮2﷯ ﷮﷮ 2𝑥 − 2﷮𝑡﷯﷯ . 𝑑𝑡﷮2𝑥 − 2﷯ I1= 1﷮2﷯ ﷮﷮ 1﷮𝑡﷯﷯ . 𝑑𝑡 I1= 1﷮2﷯ log﷮ 𝑡﷯﷯+𝐶1 I1= 1﷮2﷯ log﷮ 𝑥﷮2﷯−2𝑥−5﷯﷯+𝐶1 Now, taking I2 I2=4 ﷮﷮ 1﷮ 𝑥﷮2﷯ − 2𝑥 − 5﷯﷯ . 𝑑𝑥 I2=4 ﷮﷮ 1﷮ 𝑥﷮2﷯ − 2 𝑥﷯ 1﷯ − 5﷯﷯ . 𝑑𝑥 I2=4 ﷮﷮ 1﷮ 𝑥﷮2﷯ − 2 𝑥﷯ 1﷯ + 1﷯﷮2﷯ − 1﷯﷮2﷯ − 5﷯﷯ . 𝑑𝑥 I2=4 ﷮﷮ 1﷮ 𝑥 − 1﷯﷮2﷯ − 1﷯﷮2﷯ − 5﷯﷯ . I2=4 ﷮﷮ 1﷮ 𝑥 − 1﷯﷮2﷯ − 1 − 5﷯﷯ . 𝑑𝑥 I2=4 ﷮﷮ 1﷮ 𝑥 − 1﷯﷮2﷯ − 6﷯﷯ . 𝑑𝑥 I2=4 ﷮﷮ 1﷮ 𝑥 − 1﷯﷮2﷯ − ﷮6﷯ ﷯﷮2﷯﷯﷯ . 𝑑𝑥 I2= 4﷮2 ﷮6﷯﷯ log﷮ 𝑥 − 1 − ﷮6﷯﷮𝑥 − 1 + ﷮6﷯﷯﷯﷯+𝐶2 I2= 2﷮ ﷮6﷯﷯ log﷮ 𝑥 − 1 − ﷮6﷯﷮𝑥 − 1 + ﷮6﷯﷯﷯﷯+𝐶2 Putting the values of I1 and I2 in (1) ﷮﷮ 𝑥 + 3﷮ 𝑥﷮2﷯ − 2𝑥 − 5﷯﷯ . 𝑑𝑥= 1﷮2﷯ ﷮﷮ 2𝑥 − 2﷮ 𝑥﷮2﷯ − 2𝑥 − 5﷯﷯ . 𝑑𝑥+ ﷮﷮ 4﷮ 𝑥﷮2﷯ − 2𝑥 − 5﷯﷯ . 𝑑𝑥 = 1﷮2﷯ log﷮ 𝑥﷮2﷯−2𝑥−5﷯﷯+𝐶1+ 2﷮ ﷮6﷯﷯ log﷮ 𝑥 − 1 − ﷮6﷯﷮𝑥 − 1 + ﷮6﷯﷯﷯﷯+𝐶2 = 𝟏﷮𝟐﷯ 𝒍𝒐𝒈﷮ 𝒙﷮𝟐﷯−𝟐𝒙−𝟓﷯﷯+ 𝟐﷮ ﷮𝟔﷯﷯ 𝒍𝒐𝒈﷮ 𝒙 − 𝟏 − ﷮𝟔﷯﷮𝒙 − 𝟏 + ﷮𝟔﷯﷯﷯﷯+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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