Ex 7.4, 23 - Integrate 5x + 3 / root x2 + 4x + 10 - Integration by specific formulaes - Method 10

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.4, 23 5𝑥 + 3﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯ ﷮﷮ 5𝑥 + 3﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ 𝑑𝑥=5 ﷮﷮ 𝑥 + 3﷮5﷯﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ = 5﷮2﷯ ﷮﷮ 2𝑥 + 6﷮5﷯﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ = 5﷮2﷯ ﷮﷮ 2𝑥 + 4 − 4 + 6﷮5﷯﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ = 5﷮2﷯ ﷮﷮ 2𝑥 + 4﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ 𝑑𝑥+ 5﷮2﷯ ﷮﷮ − 14﷮5﷯﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ 𝑑𝑥 = 5﷮2﷯ ﷮﷮ 2𝑥 + 4﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ 𝑑𝑥+7 ﷮﷮ 𝑑𝑥﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ 𝑑𝑥 Solving 𝑰𝟏 I1= 5﷮2﷯ ﷮﷮ 2𝑥 + 4﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ . 𝑑𝑥 Let 𝑥﷮2﷯ + 4𝑥 + 10=𝑡 Diff both sides w.r.t.x 2𝑥+4+0= 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥= 𝑑𝑡﷮2𝑥 + 4﷯ Thus, our equation becomes I1= 5﷮2﷯ ﷮﷮ 2𝑥 + 4﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ . 𝑑𝑥 Putting the value of 𝑥﷮2﷯+4𝑥+10﷯=𝑡 and 𝑑𝑥= 𝑑𝑡﷮2𝑥 + 4﷯ I1= 5﷮2﷯ ﷮﷮ 2𝑥 + 4﷮ ﷮𝑡﷯﷯﷯ . 𝑑𝑥 I1= 5﷮2﷯ ﷮﷮ 2𝑥 + 4﷮ ﷮𝑡﷯﷯﷯ . 𝑑𝑡﷮2𝑥 + 4﷯ I1= 5﷮2﷯ ﷮﷮ 1﷮ ﷮𝑡﷯﷯﷯ . 𝑑𝑡 I1= 5﷮2﷯ ﷮﷮ 1﷮ 𝑡﷯﷮ 1﷮2﷯﷯﷯﷯ . 𝑑𝑡 I1= 5﷮2﷯ ﷮﷮ 𝑡﷯﷮ − 1﷮2﷯﷯﷯ . 𝑑𝑡 I1= 5﷮2﷯ 𝑡 ﷮ −1﷮2﷯ + 1﷯﷮ −1﷮2﷯ + 1﷯ +𝐶1 I1= 5﷮2﷯ 𝑡 ﷮ 1﷮2﷯ ﷯﷮ 1﷮2﷯﷯ +𝐶1 I1=5 𝑡 ﷮ 1﷮2﷯ ﷯+𝐶1 I1=5 ﷮𝑡﷯+𝐶1 I1=5 ﷮ 𝑥﷮2﷯+4𝑥+10﷯+𝐶1 Solving 𝑰𝟐 I2= ﷮﷮ 7﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ . 𝑑𝑥 I2=7 ﷮﷮ 1﷮ ﷮ 𝑥﷮2﷯ + 2 2﷯ 𝑥﷯ + 10﷯﷯﷯ . 𝑑𝑥 I2=7 ﷮﷮ 1﷮ ﷮ 𝑥﷮2﷯ + 2 2﷯ 𝑥﷯ + 2﷯﷮2﷯ − 2﷯﷮2﷯ + 10﷯﷯﷯ . 𝑑𝑥 I2=7 ﷮﷮ 1﷮ ﷮ 𝑥 + 2﷯﷮2﷯ − 2﷯﷮2﷯ + 10﷯﷯﷯ . 𝑑𝑥 I2=7 ﷮﷮ 1﷮ ﷮ 𝑥 + 2﷯﷮2﷯ − 4 + 10﷯﷯﷯ . 𝑑𝑥 I2=7 ﷮﷮ 1﷮ ﷮ 𝑥 + 2﷯﷮2﷯ + 6﷯﷯﷯ . 𝑑𝑥 I2=7 ﷮﷮ 1﷮ ﷮ 𝑥 + 2﷯﷮2﷯ + ﷮6﷯ ﷯﷮2﷯﷯﷯﷯ . 𝑑𝑥 I2=7 𝑙𝑜𝑔﷮ 𝑥+2+ ﷮ 𝑥+2﷯﷮2﷯ + ﷮6﷯﷯﷮2﷯﷯﷯﷯﷯+𝐶2 I2=7 𝑙𝑜𝑔﷮ 𝑥+2+ ﷮ 𝑥﷮2﷯+4𝑥+4+6﷯﷯﷯+𝐶2 I2=7 𝑙𝑜𝑔﷮ 𝑥+2+ ﷮ 𝑥﷮2﷯+4𝑥+10﷯﷯﷯+𝐶2 Putting the values of I1 and I2 in (1) ﷮﷮ 5𝑥 + 3﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ . 𝑑𝑥 = 5﷮2﷯ ﷮﷮ 2𝑥 + 4﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ . 𝑑𝑥− ﷮﷮ 7﷮ ﷮ 𝑥﷮2﷯ + 4𝑥 + 10﷯﷯﷯ . 𝑑𝑥 =5 ﷮ 𝑥﷮2﷯+4𝑥+10﷯+𝐶1−7 𝑙𝑜𝑔﷮ 𝑥+2+ ﷮ 𝑥﷮2﷯+4𝑥+10﷯﷯﷯+𝐶2 =𝟓 ﷮ 𝒙﷮𝟐﷯+𝟒𝒙+𝟏𝟎﷯−𝟕 𝒍𝒐𝒈﷮ 𝒙+𝟐+ ﷮ 𝒙﷮𝟐﷯+𝟒𝒙+𝟏𝟎﷯﷯﷯+𝑪

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