# Ex 7.5, 8

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 7.5, 8 𝑥 𝑥 − 12 (𝑥 + 2) We can write the integrand as 𝑥 𝑥 − 12 (𝑥 + 2) = 𝐴 𝑥 − 1 + 𝐵 𝑥 − 12 + 𝐶 𝑥 + 2 𝑥 𝑥 − 12 (𝑥 + 2) = 𝐴 𝑥 − 1 𝑥 + 2 + 𝐵 𝑥 + 2 + 𝐶 𝑥 −12 𝑥 − 1 𝑥 + 1 2𝑥 + 3 By cancelling denominator 𝑥 = 𝐴 𝑥−1 𝑥+2 + 𝐵 𝑥+2 + 𝐶 𝑥−12 Putting x = −2, in (1) −2 = A −2−1 −2+2 + B −2+2 + C −2−12 −2 = A×0+B×0+C −32 −2 = 9C C = − 29 Putting x = 1, in (1) 1 = A 1−1 1+2 + B 1+2 + C 1−12 1 = A×0 + B×3 + C 1−12 1 = 3B B = 13 Putting x = 0, in (1) 0=𝐴 0−1 0+2+𝐵 0+2 + 𝐶 0−12 0=𝐴 −1 2+2𝐵+𝐶 0=−2𝐴+ 23 − 29 0=−2𝐴+ 49 2𝐴= 49 ⇒𝐴= 418 ⇒𝐴= 29 Hence we can write 𝑥 𝑥 − 12 (𝑥 + 2) 𝑑𝑥 = 29 𝑥 − 1 𝑑𝑥 + 13 𝑥 − 12 𝑑𝑥 + − 29 𝑥 + 2 𝑑𝑥 = 29 𝑑𝑥 𝑥 − 1+ 13 𝑑𝑥 𝑥 − 12+ − 29 𝑑𝑥 𝑥 + 2 = 29 𝑑𝑥 𝑥 − 1 + 13 𝑑𝑥 𝑥 − 12 + − 29 𝑑𝑥 𝑥 + 2 = 29 log 𝑥−1+ 13 −1 𝑥 − 1− 29 log 𝑥+2+𝐶 = 29 log 𝑥−1− 29 log 𝑥+2− 13 𝑥 − 1 +𝐶 = 29 log 𝑥−1− log 𝑥+2− 13 𝑥 − 1 +𝐶 = 𝟐𝟗 𝐥𝐨𝐠 𝒙 − 𝟏𝒙 + 𝟐− 𝟏𝟑 𝒙 − 𝟏 +𝑪

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.