Check sibling questions
Chapter 7 Class 12 Integrals
Concept wise

Example 10 (i) - Find the following integrals (i) x + 2 / 2x2 + 6x + 5 dx - Integration by specific formulaes - Method 9

Example 10 (i) - Chapter 7 Class 12 Integrals - Part 2
Example 10 (i) - Chapter 7 Class 12 Integrals - Part 3
Example 10 (i) - Chapter 7 Class 12 Integrals - Part 4
Example 10 (i) - Chapter 7 Class 12 Integrals - Part 5
Example 10 (i) - Chapter 7 Class 12 Integrals - Part 6


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Transcript

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

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Davneet Singh

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.