Example 10 (i) - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 10 Find the following integrals: (i) 𝑥 + 22 𝑥2 + 6𝑥 + 5 𝑑𝑥 It can be written as 𝑥+2= A 𝑑𝑑𝑥 2 𝑥2+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= 14 4𝑥+6+ 12 Now, our equation is 𝑥 + 22 𝑥2 + 6𝑥 + 5.𝑑𝑥= 14 4𝑥 + 6 + 122 𝑥2 + 6𝑥 + 5.𝑑𝑥 = 14 4𝑥 + 62 𝑥2 + 6𝑥 + 5+ 122 𝑥2+6𝑥+5.𝑑𝑥 = 14 4𝑥 + 62 𝑥2 + 6𝑥 + 5𝑑𝑥+ 12 12 𝑥2 + 6𝑥 + 5.𝑑𝑥 Taking I1 I1 = 14 4𝑥 + 62 𝑥2 + 6𝑥 + 5𝑑𝑥 Let t = 2 𝑥2 + 6𝑥 + 5 Differentiating both sides w.r.t.𝑥 4𝑥 +6= 𝑑𝑡𝑑𝑥 𝑑𝑥= 𝑑𝑡4𝑥 + 6 Now, I1 = 14 4𝑥 + 62 𝑥2 + 6𝑥 + 5.𝑑𝑥 Putting the values of 2 𝑥2+6𝑥+5 and 𝑑𝑥, we get I1 = 14 4𝑥 + 6𝑡. 𝑑𝑡4𝑥 + 6 I1 = 14 1𝑡.𝑑𝑡 I1 = 14𝑙𝑜𝑔 𝑡+𝐶1 I1 = 14𝑙𝑜𝑔 2 𝑥2+6𝑥+5+𝐶1 Now, taking I2 i.e. I2 = 12 12 𝑥2 + 6𝑥 + 5.𝑑𝑥 I2 = 12 12 𝑥2 + 6𝑥2 + 52 .𝑑𝑥 I2 = 12.2 1 𝑥2 +3𝑥 + 52.𝑑𝑥 I2 = 14 1 𝑥2 + 2 𝑥 32 + 52.𝑑𝑥 I2 = 14 1 𝑥2 + 2 𝑥 32 + 322 − 322 + 52.𝑑𝑥 I2 = 14 1 𝑥 + 322 − 322 + 52𝑑𝑥 I2 = 14 1 𝑥 + 322 − 94 + 52.𝑑𝑥 I2 = 14 1 𝑥 + 322+ −9 + 104 .𝑑𝑥 I2 = 14 1 𝑥 + 322+ 14 .𝑑𝑥 I2 = 14 1 𝑥 + 322+ 122 .𝑑𝑥 = 14 1 12 tan−1 𝑥 + 32 12+𝐶2 = 14 2 tan−1 2𝑥 + 32 12+𝐶2 = 14 2 tan−1 2𝑥+3+𝐶2 = 24 tan−1 2𝑥+3+ 𝐶24 = 12 tan−1 2𝑥+3+𝐶3 Now, putting the value of I1 and I2 in eq. (1) ∴ 𝑥+22 𝑥2 + 6𝑥 + 5.𝑑𝑥 = 14𝑙𝑜𝑔 2 𝑥2+6𝑥+5+𝐶1+ 12 tan−1 2𝑥+3+𝐶3 = 𝟏𝟒𝒍𝒐𝒈 𝟐 𝒙𝟐+𝟔𝒙+𝟓+ 𝟏𝟐 𝒕𝒂𝒏−𝟏 𝟐𝒙+𝟑+𝑪
Integration by specific formulaes - Method 9
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo