Check sibling questions

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

Ex 7.4, 7 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.4, 7 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.4, 7 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.4, 7 - Chapter 7 Class 12 Integrals - Part 5

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Transcript

Ex 7.4, 7 𝑥 − 1﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯ Integrating the function 𝑤.𝑟.𝑡.𝑥 ﷮﷮ 𝑥 − 1﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ 𝑑𝑥 = ﷮﷮ 𝑥﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯ − 1﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯﷯ 𝑑𝑥 = ﷮﷮ 𝑥﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ 𝑑𝑥− ﷮﷮ 1﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ 𝑑𝑥 Solving 𝐈𝟏 I1 = ﷮﷮ 𝑥﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ 𝑑𝑥 Let 𝑥﷮2﷯ − 1=𝑡 Diff both sides w.r.t.x 2𝑥−0= 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥= 𝑑𝑡﷮2𝑥﷯ Thus, our equation becomes ∴ I1 = ﷮﷮ 𝑥﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ 𝑑𝑥 Put the values of 𝑥﷮2﷯ −1﷯=𝑡 and 𝑑𝑥= 𝑑𝑡﷮2𝑥﷯ I1 = ﷮﷮ 𝑥﷮ ﷮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 = ﷮﷮ 1﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ 𝑑𝑥 I2 = ﷮﷮ 1﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷮2﷯﷯﷯﷯ 𝑑𝑥 I2 = log﷮ 𝑥+ ﷮ 𝑥﷮2﷯ − 1﷯﷮2﷯﷯﷯﷯+𝐶2 I2 = log﷮ 𝑥+ ﷮ 𝑥﷮2﷯ −1﷯ ﷯﷯+𝐶2 Now, putting the value of I1 and I2 in eq. (1) ﷮﷮ 𝑥 − 1﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ 𝑑𝑥= ﷮﷮ 𝑥﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ 𝑑𝑥− ﷮﷮ 1﷮ ﷮ 𝑥﷮2﷯ − 1﷯﷯﷯ 𝑑𝑥 = ﷮ 𝑥﷮2﷯−1﷯ +𝐶1− log﷮ 𝑥+ ﷮ 𝑥﷮2﷯ −1﷯ ﷯﷯+𝐶2 ﷯ = ﷮ 𝑥﷮2﷯−1﷯ +𝐶1− log﷮ 𝑥+ ﷮ 𝑥﷮2﷯ −1﷯ ﷯﷯−𝐶2 = ﷮ 𝒙﷮𝟐﷯−𝟏﷯ − 𝒍𝒐𝒈﷮ 𝒙+ ﷮ 𝒙﷮𝟐﷯ −𝟏﷯ ﷯﷯−𝑪

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.