Example 10 (i) - Chapter 7 Class 12 Integrals (Term 2)
Last updated at July 9, 2018 by Teachoo
Last updated at July 9, 2018 by Teachoo
Transcript
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
Integration by specific formulaes - Method 9
About the Author