# Misc 6 - Chapter 7 Class 12 Integrals (Term 2)

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 6 Integrate the function 5𝑥 𝑥 + 1 𝑥2 + 9 Let I = 5𝑥 𝑥 + 1 𝑥2 + 9𝑑𝑥 We can write integrate as : 5𝑥 𝑥 + 1 𝑥2+ 9= A𝑥 + 1+ B𝑥 + 𝐶 𝑥2 + 9 By cancelling denominators 5𝑥=A 𝑥2+9+ B𝑥+𝐶 𝑥+1 Putting 𝑥=−1 5 × −1=A −12+9+ B × −1+𝐶 −1+1 – 5 = A 1+9+ −B+𝐶 0 – 5 = A × 10 – 5 = 10A A = −5 10 A = −1 2 Similarly putting 𝑥=0 5 × 0 = A 02+9+ B ×0+𝐶 0+1 0 = A 9+ 0+𝐶 1 0 = 9A + C C = – 9A C = – 9 × −1 2 C = 92 Putting x = 1 5(1)= A 12+9+ B1+𝐶 1+1 5(1)= A(10) +(B + C) (2) Putting A = −12 and C = 92 5(1) = −12 (10) + 2B + 2C 5(1) = −5 + 2B + 2 92 5(1) = −5 + 2B + 9 5 = 2B + 4 1 = 2B 12 = B Hence we can write 5𝑥 𝑥+1 𝑥2+9= −1 2𝑥 + 1+ 12 𝑥 + 92 𝑥2 + 9 = −12 𝑥 + 1+ 𝑥2 𝑥2 + 9+ 92 𝑥2 + 9 Integrating w.r.t.𝑥 5𝑥 𝑥 + 1 𝑥2 + 9𝑑𝑥= −12 𝑥 + 1+ 𝑥2 𝑥2 + 9+ 92 𝑥2 + 9 = −12 𝑥 + 1𝑑𝑥+ 𝑥2 𝑥2 + 9𝑑𝑥+ 92 𝑥2 + 9𝑑𝑥 Hence I = I1 + I2 + I3 I1 = −12 𝑥 + 1𝑑𝑥 = −1 2 𝑙𝑜𝑔 𝑥+1+𝐶1 I2 = 𝑥2 𝑥2 + 9𝑑𝑥 = 12 𝑥 𝑥2 + 9𝑑𝑥 Putting 𝑡= 𝑥2+9 Differentiating w.r.t. 𝑥 𝑑𝑡𝑑𝑥=2𝑥 𝑑𝑡2𝑥=𝑑𝑥 Therefore 12 𝑥 𝑥2 + 9𝑑𝑥= 12 𝑥𝑡 × 𝑑𝑡2𝑥 = 12 𝑑𝑡2𝑡 = 12 ×2 𝑑𝑡𝑡 = 14log 𝑡+𝐶2 Putting back 𝑡= 𝑥2+9 = 14𝑙𝑜𝑔 𝑥2+9+𝐶2 Now, I3 = 92 𝑥2 + 9𝑑𝑥 = 92 1 𝑥2 + 9𝑑𝑥 Now, I3 = 92 𝑥2 + 9𝑑𝑥 = 92 1 𝑥2 + 9𝑑𝑥 = 92 1 𝑥2 + 32𝑑𝑥 = 92 × 13 tan−1 𝑥3+𝐶3 = 32 tan−1 𝑥3+ 𝐶3 Hence I = I1 + I2 + I3 = −1 2𝑙𝑜𝑔 𝑥+1+𝐶1+ 14𝑙𝑜𝑔 𝑥2+9+𝐶2+ 32 tan−1 𝑥3+𝐶3 = −1 2𝑙𝑜𝑔 𝑥+1+ 14𝑙𝑜𝑔 𝑥2+9+ 32 tan−1 𝑥3+𝐶1+𝐶2+𝐶3 = −𝟏 𝟐𝒍𝒐𝒈 𝒙+𝟏+ 𝟏𝟒𝒍𝒐𝒈 𝒙𝟐+𝟗+ 𝟑𝟐 𝐭𝐚𝐧−𝟏 𝒙𝟑+𝑪

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Integration Formula Sheet - Chapter 7 Class 12 Formulas Important

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.