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Ex 7.5

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Ex 7.5, 5 You are here

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Ex 7.5, 22 (MCQ)

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Last updated at May 29, 2023 by Teachoo

Ex 7.5, 5 2𝑥 𝑥2 + 3𝑥 + 2 Solving integrand 2𝑥 𝑥2 + 3𝑥 + 2 = 2𝑥 𝑥2 + 2𝑥 + 𝑥 + 2 = 2𝑥𝑥 𝑥 + 2 + 1 𝑥 + 2 = 2𝑥 𝑥 + 2 𝑥 + 1 We can write it as 2𝑥 𝑥 + 2 𝑥 + 1 = 𝐴 𝑥 + 2 + 𝐵 𝑥 + 1 2𝑥 𝑥 + 2 𝑥 + 1 = 𝐴 𝑥 + 1 + 𝐵 𝑥 + 2 𝑥 + 2 𝑥 + 1 Cancelling denominator 2𝑥 = 𝐴 𝑥+1+𝐵 𝑥+2 Putting x = −1, in (1) 2 −1 = 𝐴 −1+1 + 𝐵 −1+2 −2 = 𝐴×0+𝐵 1 𝐵 = −2 Similarly putting x = − 2 , in (1) 2 −2 = 𝐴 −2+1 + 𝐵 −2+2 −4 = 𝐴 −1+𝐵×0 −4 = −𝐴 𝐴 = 4 Hence we can write it as 2𝑥 𝑥 + 2 𝑥 + 1 = 4 𝑥 + 2 + −2 𝑥 + 1 = 4 𝑥 + 2 − 2 𝑥 + 1 Integrating 𝑤.𝑟.𝑡.𝑥. 2𝑥 𝑥 + 2 𝑥 + 1 𝑑𝑥 = 4 𝑥 + 2 − 2 𝑥 + 1𝑑𝑥 = 4 𝑑𝑥 𝑥 + 2−2 𝑑𝑥 𝑥 + 1 = 𝟒 𝐥𝐨𝐠 𝒙+𝟐−𝟐 𝐥𝐨𝐠 𝒙+𝟏 +𝑪