Ex 7.5, 9 - Integrate 3x + 5 / x3 - x2 - x + 1 - Ex 7.5

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.5, 9 3𝑥 + 5﷮ 𝑥﷮3﷯ − 𝑥﷮2﷯ − 𝑥 + 1﷯ Let I= ﷮﷮ 3𝑥 + 5﷮ 𝑥﷮3﷯ − 𝑥﷮2﷯ − 𝑥 + 1﷯﷯ 𝑑𝑥 We can write integrand as 3𝑥 + 5﷮ 𝑥﷮3﷯ − 𝑥﷮2﷯ − 𝑥 + 1﷯= 3𝑥 + 5﷮ 𝑥 − 1﷯ 𝑥﷮2﷯ − 1﷯﷯ = 3𝑥 + 5﷮ 𝑥 − 1﷯ 𝑥 − 1﷯ 𝑥 + 1﷯﷯ = 3𝑥 + 5﷮ 𝑥 + 1﷯ 𝑥 − 1﷯﷮2﷯﷯ We can write it as 3𝑥 + 5﷮ 𝑥 + 1﷯ 𝑥 − 1﷯﷮2﷯﷯= 𝐴﷮ 𝑥 + 1﷯﷯ + 𝐵﷮ 𝑥 − 1﷯﷯ + 𝐶﷮ 𝑥 − 1﷯﷮2﷯﷯ 3𝑥 + 5﷮ 𝑥 + 1﷯ 𝑥 − 1﷯﷮2﷯﷯= 𝐴 𝑥 − 1﷯﷮2﷯ + 𝐵 𝑥 + 1﷯ 𝑥 − 1﷯ + 𝐶 𝑥 + 1﷯﷮ 𝑥 + 1﷯ 𝑥 − 1﷯ 𝑥 − 1﷯﷯ 3𝑥 + 5﷮ 𝑥 + 1﷯ 𝑥 − 1﷯﷮2﷯﷯= 𝐴 𝑥 − 1﷯﷮2﷯ + 𝐵 𝑥﷮2﷯ − 1﷯ + 𝐶 𝑥 + 1﷯﷮ 𝑥 + 1﷯ 𝑥 − 1﷯﷮2﷯﷯ By Cancelling denominator 3𝑥+5=𝐴 𝑥−1﷯﷮2﷯+𝐵 𝑥﷮2﷯−1﷯+𝐶 𝑥+1﷯ Put 𝑥=1 , in (1) 3×1+5=𝐴 1−1﷯﷮2﷯+𝐵 1﷮2﷯−1﷯+𝐶 1+1﷯ 8=𝐴×0+ 𝐵×0+𝐶×2 8=2𝐶 𝐶=4 Similarly put 𝑥=−1 , in (1) 3𝑥+5=𝐴 𝑥−1﷯﷮2﷯+𝐵 𝑥﷮2﷯−1﷯+𝐶 𝑥+1﷯ 3 −1﷯+5=𝐴 −1−1﷯﷮2﷯+𝐵 −1﷯﷮2﷯−1﷯+𝐶 −1+1﷯ −3+5=𝐴 −2﷯﷮2﷯+𝐵 1−1﷯+𝐶 0﷯ 2=4𝐴+𝐵×0+𝐶×0 2=4𝐴 𝐴= 1﷮2﷯ Putting x = 0 , in (1) 3 0﷯+5=𝐴 0−1﷯﷮2﷯+𝐵 0−1﷯+𝐶 0+1﷯ 5=𝐴−𝐵+𝐶 5= 1﷮2﷯ −𝐵+4 5= −𝐵+ 9﷮2﷯ 𝐵= 9﷮2﷯ −5 𝐵=− 1﷮2﷯ Hence we can write 3𝑥 + 5﷮ 𝑥 + 1﷯ 𝑥 − 1﷯﷮2﷯﷯= 1﷮2﷯﷯﷮ 𝑥 + 1﷯﷯ + − 1﷮2﷯﷮ 𝑥 − 1﷯﷯ + 4﷮ 𝑥 − 1﷯﷮2﷯﷯ 3𝑥 + 5﷮ 𝑥 + 1﷯ 𝑥 − 1﷯﷮2﷯﷯= 1﷮2 𝑥 + 1﷯﷯ − 1﷮2 𝑥 − 1﷯﷯ + 4﷮ 𝑥 − 1﷯﷮2﷯﷯ Integrating 𝑤.𝑟.𝑡.𝑥 I= ﷮﷮ 3𝑥 + 5﷮ 𝑥﷮3﷯ − 𝑥﷮2﷯ − 𝑥 + 1﷯﷯ 𝑑𝑥= ﷮﷮ 1﷮2 𝑥 + 1﷯﷯ − 1﷮2 𝑥 − 1﷯﷯ + 4﷮ 𝑥 − 1﷯﷮2﷯﷯﷯﷯ 𝑑𝑥 = 1﷮2﷯ ﷮﷮ 𝑑𝑥﷮𝑥 + 1﷯﷯ − 1﷮2﷯ ﷮﷮ 𝑑𝑥﷮ 𝑥 − 1﷯﷯﷯+4 ﷮﷮ 𝑑𝑥﷮ 𝑥 − 1﷯﷮2﷯﷯﷯ Hence I=I1−I2+I3 Now, I1= 1﷮2﷯ ﷮﷮ 1﷮𝑥 + 1﷯﷯ 𝑑𝑥 = 1﷮2﷯ log﷮ 𝑥+1﷯﷯+𝐶1 Also, I2 = 1﷮2﷯ ﷮﷮ 1﷮𝑥 − 1﷯﷯ 𝑑𝑥 = 1﷮2﷯ log﷮ 𝑥−1﷯﷯+𝐶2 And, I3= ﷮﷮ 4﷮ 𝑥 − 1﷯﷮2﷯﷯﷯ 𝑑𝑥 =4 ﷮﷮ 1﷮ 𝑥 − 1﷯﷮2﷯﷯﷯ 𝑑𝑥 =4 ﷮﷮ 𝑥 − 1﷯﷮−2﷯﷯ 𝑑𝑥 = 4 𝑥 − 1﷯﷮−2 + 1﷯﷮−2 + 1﷯ +𝐶3 = 4 𝑥 − 1﷯﷮−1﷯﷮−1﷯ +𝐶3 = − 4﷮𝑥 − 1﷯+𝐶3 Therefore I=I1−I2+I3 I= 1﷮2﷯ log﷮ 𝑥+1﷯﷯+𝐶1− 1﷮2﷯ log﷮ 𝑥−1﷯﷯−𝐶2+ − 4﷮𝑥 − 1﷯+𝐶3 = 1﷮2﷯ log﷮ 𝑥+1﷯﷯+− 1﷮2﷯ log﷮ 𝑥−1﷯﷯− 4﷮𝑥 − 1﷯ +𝐶1−𝐶2+𝐶3 = 1﷮2﷯ log﷮ 𝑥+1﷯﷯− log﷮ 𝑥−1﷯﷯﷯− 4﷮𝑥 − 1﷯ +𝐶 = 𝟏﷮𝟐﷯ 𝒍𝒐𝒈﷮ 𝒙 + 𝟏﷮𝒙−𝟏﷯﷯﷯− 𝟒﷮𝒙 − 𝟏﷯ +𝑪

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