# Ex 7.5, 18

Last updated at Sept. 29, 2017 by Teachoo

Last updated at Sept. 29, 2017 by Teachoo

Transcript

Ex 7.5, 18 𝑥2+ 1 𝑥2+ 2 𝑥2+ 3 𝑥2+ 4 Let t = 𝑥2 𝑥2 + 1 𝑥2 + 2 𝑥2 + 3 𝑥2 + 4 = 𝑡 + 1 𝑡 + 2 𝑡 + 3 𝑡 + 4 = 𝑡2 + 3𝑡 + 2 𝑡2 + 7𝑡 + 12 = 1 + −4𝑡 −10 𝑡2 + 7𝑡 + 12 = 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 By 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 −6 = 𝐴×0+𝐵 −1 −6 = −𝐵 𝐵 = 6 Similarly putting t = −3 in (1) 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 = 𝑥2 4 𝑥2 − 10 𝑥2 + 3 𝑥2 + 4 = −2 𝑥2 + 3 + 6 𝑥2 + 4 Therefore 𝑥2+ 1 𝑥2+ 2 𝑥2+ 3 𝑥2+ 4 = 1− −2 𝑥2 + 3 + 6 𝑥2 + 4 𝑑𝑥 = 1. 𝑑𝑥 + 2 𝑥2 + 3 𝑑𝑥 − 6 𝑥2 + 4 𝑑𝑥 = 1. 𝑑𝑥 + 2 1 𝑥2 + 32 𝑑𝑥 − 6 1 𝑥2 + 22 𝑑𝑥 = 𝑥 + 2 × 1 3 tan−1 𝑥 3 − 6 × 12 tan−1 𝑥2+𝐶 = 𝒙 + 𝟐 𝟑 𝒕𝒂𝒏−𝟏 𝒙 𝟑−𝟑 𝒕𝒂𝒏−𝟏 𝒙𝟐+𝑪

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.