Ex 7.5, 18 - Integrate (x2 + 1) (x2 + 2) / (x3+3) (x2+4)

Ex 7.5, 18 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.5, 18 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.5, 18 - Chapter 7 Class 12 Integrals - Part 4

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Ex 7.5, 18 Integrate the function (𝑥2+ 1)(𝑥2+ 2)/(𝑥2+ 3)(𝑥2+ 4) (𝑥^2 + 1)(𝑥^2 + 2)/(𝑥^2 + 3)(𝑥^2 + 4) " " Let t = 𝑥^2 = (𝑡 + 1)(𝑡 + 2)/(𝑡 + 3)(𝑡 + 4) = (𝑡^2 + 3𝑡 + 2)/(𝑡^2 + 7𝑡 + 12) = 1 + (−4𝑡 −10)/(𝑡^2 + 7𝑡 + 12) = 1 + (−(4𝑡 + 10))/(𝑡 + 3)(𝑡 + 4) Rough = 1 − ( (4𝑡 + 10))/(𝑡 + 3)(𝑡 + 4) We can write (4𝑡 + 10)/((𝑡 + 3) (𝑡 + 4) ) = 𝐴/((𝑡 + 3) ) + 𝐵/((𝑡 + 4) ) (4𝑡 + 10)/((𝑡 + 3) (𝑡 + 4) ) = (𝐴(𝑡 + 4) + 𝐵(𝑡 + 3))/((𝑡 + 3) (𝑡 + 4) ) Cancelling denominator 4𝑡−10 = 𝐴(𝑡+4)+𝐵(𝑡+3) Putting t = − 4 in (1) 4(−4)+10 = 𝐴(−4+4)+𝐵(−4+3) −16+10 = 𝐴×0+𝐵(−1) …(1) −6 = 𝐴×0+𝐵(−1) −6 = −𝐵 𝐵 = 6 Putting t = −3 in (1) 4𝑡−10 = 𝐴(𝑡+4)+𝐵(𝑡+3) 4(−3)+10 = 𝐴(−3+4)+𝐵(−3+3) −12+10 = 𝐴×1+𝐵×0 −2 = 𝐴 𝐴 = −2 Hence we can write (4𝑡 + 10)/((𝑡 + 3) (𝑡 + 4) ) = (−2)/((𝑡 + 3) ) + 6/((𝑡 + 4) ) Putting back t = 𝑥^2 (4𝑥^2 − 10)/((𝑥^2 + 3) (𝑥^2 + 4) ) = (−2)/((𝑥^2 + 3) ) + 6/((𝑥^2 + 4) ) Therefore ∫1▒(𝑥2+ 1)(𝑥2+ 2)/(𝑥2+ 3)(𝑥2+ 4) = ∫1▒〖1−[(−2)/((𝑥^2 + 3) ) + 6/((𝑥^2 + 4) )] 〗 𝑑𝑥 = ∫1▒1. 𝑑𝑥 + ∫1▒2/((𝑥^2 + 3) ) 𝑑𝑥 − ∫1▒6/((𝑥^2 + 4) ) 𝑑𝑥 = ∫1▒1. 𝑑𝑥 + 2∫1▒1/(𝑥^2 + (√3)^2 ) 𝑑𝑥 − 6∫1▒1/((𝑥^2 +2^2 ) ) 𝑑𝑥 = 𝑥 + 2 × 1/√3 tan^(−1)⁡〖 𝑥/√3〗 − 6 × 1/2 tan^(−1)⁡〖 𝑥/2〗+𝐶 = 𝒙 + 𝟐/√𝟑 〖𝒕𝒂𝒏〗^(−𝟏)⁡(𝒙/√𝟑)−𝟑 〖𝒕𝒂𝒏〗^(−𝟏)⁡(𝒙/𝟐)+𝑪

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.