Example 10 - Find integrals (i) x + 2 / 2x2 + 6x + 5 dx - Examples

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Example 10 Find the following integrals: (i) ﷮﷮ 𝑥 + 2﷮2 𝑥﷮2﷯ + 6𝑥 + 5 ﷯﷯ 𝑑𝑥 It can be written as 𝑥+2= A 𝑑﷮𝑑𝑥﷯ 2 𝑥﷮2﷯+6𝑥+5﷯+ B 𝑥+2= A 4𝑥+6﷯+ B 𝑥+2= 4A﷯ 𝑥﷯+6A+B Finding A & B Now, we know that 𝑥+2= A 4𝑥+6﷯+ B 𝑥+2= 1﷮4﷯ 4𝑥+6﷯+ 1﷮2﷯ Now, our equation is ﷮﷮ 𝑥 + 2﷮2 𝑥﷮2﷯ + 6𝑥 + 5﷯.𝑑𝑥= ﷮﷮ 1﷮4﷯ 4𝑥 + 6﷯ + 1﷮2﷯﷮2 𝑥﷮2﷯ + 6𝑥 + 5﷯.𝑑𝑥﷯﷯ = ﷮﷮ 1﷮4﷯ 4𝑥 + 6﷯﷮2 𝑥﷮2﷯ + 6𝑥 + 5﷯+ ﷮﷮ 1﷮2﷯﷮2 𝑥﷮2﷯+6𝑥+5﷯.𝑑𝑥﷯﷯ = 1﷮4﷯ ﷮﷮ 4𝑥 + 6﷮2 𝑥﷮2﷯ + 6𝑥 + 5﷯𝑑𝑥+ 1﷮2﷯ ﷮﷮ 1﷮2 𝑥﷮2﷯ + 6𝑥 + 5﷯.𝑑𝑥﷯﷯ Taking I1 I1 = 1﷮4﷯ ﷮﷮ 4𝑥 + 6﷮2 𝑥﷮2﷯ + 6𝑥 + 5﷯𝑑𝑥﷯ Let t = 2 𝑥﷮2﷯ + 6𝑥 + 5 Differentiating both sides w.r.t.𝑥 4𝑥 +6= 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥= 𝑑𝑡﷮4𝑥 + 6﷯ Now, I1 = 1﷮4﷯ ﷮﷮ 4𝑥 + 6﷮2 𝑥﷮2﷯ + 6𝑥 + 5﷯.𝑑𝑥﷯ Putting the values of 2 𝑥﷮2﷯+6𝑥+5﷯ and 𝑑𝑥, we get I1 = 1﷮4﷯ ﷮﷮ 4𝑥 + 6﷮𝑡﷯. 𝑑𝑡﷮4𝑥 + 6﷯ ﷯ I1 = 1﷮4﷯ ﷮﷮ 1﷮𝑡﷯.𝑑𝑡 ﷯ I1 = 1﷮4﷯𝑙𝑜𝑔 𝑡﷯+𝐶1 I1 = 1﷮4﷯𝑙𝑜𝑔 2 𝑥﷮2﷯+6𝑥+5﷯+𝐶1 Now, taking I2 i.e. I2 = 1﷮2﷯ ﷮﷮ 1﷮2 𝑥﷮2﷯ + 6𝑥 + 5﷯.𝑑𝑥 ﷯ I2 = 1﷮2﷯ ﷮﷮ 1﷮2 𝑥﷮2﷯ + 6𝑥﷮2﷯ + 5﷮2﷯ ﷯﷯.𝑑𝑥 ﷯ I2 = 1﷮2.2﷯ ﷮﷮ 1﷮ 𝑥﷮2﷯ +3𝑥 + 5﷮2﷯﷯.𝑑𝑥 ﷯ I2 = 1﷮4﷯ ﷮﷮ 1﷮ 𝑥﷮2﷯ + 2 𝑥﷯ 3﷮2﷯﷯ + 5﷮2﷯﷯.𝑑𝑥 ﷯ I2 = 1﷮4﷯ ﷮﷮ 1﷮ 𝑥﷮2﷯ + 2 𝑥﷯ 3﷮2﷯﷯ + 3﷮2﷯﷯﷮2﷯ − 3﷮2﷯﷯﷮2﷯ + 5﷮2﷯﷯.𝑑𝑥 ﷯ I2 = 1﷮4﷯ ﷮﷮ 1﷮ 𝑥 + 3﷮2﷯﷯﷮2﷯ − 3﷮2﷯﷯﷮2﷯ + 5﷮2﷯﷯𝑑𝑥 ﷯ I2 = 1﷮4﷯ ﷮﷮ 1﷮ 𝑥 + 3﷮2﷯﷯﷮2﷯ − 9﷮4﷯ + 5﷮2﷯﷯.𝑑𝑥 ﷯ I2 = 1﷮4﷯ ﷮﷮ 1﷮ 𝑥 + 3﷮2﷯﷯﷮2﷯+ −9 + 10﷮4﷯ ﷯.𝑑𝑥 ﷯ I2 = 1﷮4﷯ ﷮﷮ 1﷮ 𝑥 + 3﷮2﷯﷯﷮2﷯+ 1﷮4﷯ ﷯.𝑑𝑥 ﷯ I2 = 1﷮4﷯ ﷮﷮ 1﷮ 𝑥 + 3﷮2﷯﷯﷮2﷯+ 1﷮2﷯﷯﷮2﷯ ﷯.𝑑𝑥 ﷯ = 1﷮4﷯ 1﷮ 1﷮2﷯﷯ tan﷮−1﷯﷮ 𝑥 + 3﷮2﷯﷮ 1﷮2﷯﷯+𝐶2﷯﷯ = 1﷮4﷯ 2 tan﷮−1﷯﷮ 2𝑥 + 3﷮2﷯﷮ 1﷮2﷯﷯+𝐶2﷯﷯ = 1﷮4﷯ 2 tan﷮−1﷯ 2𝑥+3﷯﷮+𝐶2﷯﷯ = 2﷮4﷯ tan﷮−1﷯﷮ 2𝑥+3﷯+ 𝐶2﷮4﷯﷯ = 1﷮2﷯ tan﷮−1﷯﷮ 2𝑥+3﷯+𝐶3﷯ Now, putting the value of I1 and I2 in eq. (1) ∴ ﷮﷮ 𝑥+2﷮2 𝑥﷮2﷯ + 6𝑥 + 5﷯.𝑑𝑥﷯ = 1﷮4﷯𝑙𝑜𝑔 2 𝑥﷮2﷯+6𝑥+5﷯+𝐶1+ 1﷮2﷯ tan﷮−1﷯﷮ 2𝑥+3﷯+﷯𝐶3 = 𝟏﷮𝟒﷯𝒍𝒐𝒈 𝟐 𝒙﷮𝟐﷯+𝟔𝒙+𝟓﷯+ 𝟏﷮𝟐﷯ 𝒕𝒂𝒏﷮−𝟏﷯﷮ 𝟐𝒙+𝟑﷯+﷯𝑪 Example 10 Find the following integrals: (ii) ﷮﷮ 𝑥 + 3﷮ ﷮5 − 4𝑥 − 𝑥﷮2﷯﷯﷯﷯ 𝑑𝑥 It can be written in the form 𝑥+3= A 𝑑﷮𝑑𝑥﷯ − 𝑥﷮2﷯−4𝑥+5﷯﷯+ B 𝑥+3= A −2𝑥−4﷯+ B 𝑥+3= 2A﷯ 𝑥﷯−4A+B Finding A & B Now, we know that 𝑥+3=A −2𝑥−4﷯+B 𝑥+3= −1﷮ 2﷯ −2𝑥−4﷯+1 Step 2 : Integrating ﷮﷮ 𝑥+3﷮ ﷮5 − 4𝑥 + 𝑥﷮2﷯﷯﷯﷯.𝑑𝑥= ﷮﷮ 1﷮2﷯ −2𝑥 − 4﷯+1﷮ ﷮5 − 4𝑥 − 𝑥﷮2﷯﷯﷯﷯.𝑑𝑥 = ﷮﷮ −1﷮ 2﷯ −2𝑥 − 4﷯+1﷮ ﷮5 − 4𝑥 + 𝑥﷮2﷯﷯﷯﷯𝑑𝑥+ ﷮﷮ 1﷮ ﷮5 − 4𝑥 − 𝑥﷮2﷯﷯﷯﷯𝑑𝑥 = −1﷮ 2﷯ ﷮﷮ −2𝑥 − 4﷮ ﷮5 − 4𝑥 + 𝑥﷮2﷯﷯﷯﷯.𝑑𝑥+ ﷮﷮ 1﷮ ﷮5 − 4𝑥 − 𝑥﷮2﷯﷯﷯﷯.𝑑𝑥 Taking I1 I1 = −1﷮ 2﷯ ﷮﷮ −2𝑥 − 4﷮ ﷮− 𝑥﷮2﷯ − 4𝑥 + 5﷯﷯.𝑑𝑥﷯ Let t = − 𝑥﷮2﷯ − 4𝑥 + 5 Differentiating both sides w.r.t. 𝑥 −2𝑥 − 4 = 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥= 𝑑𝑡﷮− 2𝑥 − 4﷯ Now, I1 = −1﷮ 2﷯ ﷮﷮ −2𝑥 − 4﷮ ﷮− 𝑥﷮2﷯ − 4𝑥 + 5﷯﷯.𝑑𝑥﷯ Putting the values of 𝑥﷮2﷯−4𝑥+5﷯ and 𝑑𝑥 I1 = −1﷮ 2﷯ ﷮﷮ −2𝑥 − 4﷮ ﷮𝑡﷯﷯. 𝑑𝑡﷮ −2𝑥 −4﷯﷯﷯ 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﷯−4𝑥+5﷯+𝐶1 I1 = − ﷮5−4𝑥− 𝑥﷮2﷯﷯+𝐶1 Taking 𝐈𝟐 I2 = ﷮﷮ 1﷮ ﷮5 − 4𝑥 − 𝑥﷮2﷯﷯﷯.𝑑𝑥﷯ I2 = ﷮﷮ 1﷮ ﷮− 𝑥﷮2﷯ + 4𝑥 − 5﷯﷯﷯𝑑𝑥﷯ I2 = ﷮﷮ 1﷮ ﷮− 𝑥﷮2﷯ + 2 2﷯ 𝑥﷯ − 5﷯﷯﷯𝑑𝑥﷯ I2 = ﷮﷮ 1﷮ ﷮− 𝑥﷮2﷯ + 2 2﷯ 𝑥﷯ + 2﷯﷮2﷯ − 2﷯﷮2﷯ − 5﷯﷯﷯𝑑𝑥﷯ I2 = ﷮﷮ 1﷮ ﷮− 𝑥+2﷯﷮2﷯ − 4 − 5﷯﷯﷯𝑑𝑥﷯ I2 = ﷮﷮ 1﷮ ﷮− 𝑥+2﷯﷮2﷯ − 9﷯﷯﷯𝑑𝑥﷯ I2 = ﷮﷮ 1﷮ ﷮9 − 𝑥 + 2﷯﷮2﷯﷯﷯𝑑𝑥﷯ I2 = ﷮﷮ 1﷮ ﷮ 3﷯﷮2﷯− 𝑥+2﷯﷮2﷯﷯﷯𝑑𝑥﷯ I2 = sin﷮−1﷯﷮ 𝑥 + 2﷮3﷯﷯+𝐶2﷯ Now, putting the values of I1 and I2 in eq. (1) we get ﷮﷮ 𝑥 + 3﷮ ﷮5 − 4𝑥 − 𝑥﷮2﷯﷯﷯.𝑑𝑥= −1﷮ 2﷯ ﷮﷮ − 2𝑥 − 4﷮ ﷮5 − 4𝑥 − 𝑥﷮2﷯﷯﷯﷯﷯+ ﷮﷮ 1﷮ ﷮5 − 4𝑥 − 𝑥﷮2﷯﷯﷯﷯𝑑𝑥 =− ﷮5 − 4𝑥 − 𝑥﷮2﷯﷯+𝐶1+ sin﷮−1﷯﷮ 𝑥 + 2﷮3﷯﷯+𝐶2﷯ =− ﷮𝟓 − 𝟒𝒙 − 𝒙﷮𝟐﷯﷯+ 𝒔𝒊𝒏﷮−𝟏﷯﷮ 𝒙 + 𝟐﷮𝟑﷯﷯+𝑪﷯

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