Ex 7.4

Chapter 7 Class 12 Integrals
Serial order wise

### 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