# Ex 7.5, 11

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 7.5, 11 5𝑥 𝑥 + 1 𝑥2− 4 We can write the integrand as 5𝑥 𝑥 + 1 𝑥2− 4 = 5𝑥 𝑥 + 1 𝑥 − 2 𝑥 + 2 5𝑥 𝑥 + 1 𝑥2− 4 = 𝐴 𝑥 + 1 + 𝐵 𝑥 − 2 + 𝐶 𝑥 + 2 5𝑥 𝑥 + 1 𝑥2− 4 = 𝐴 𝑥 − 2 𝑥 + 2 + 𝐵 𝑥 + 1 𝑥 + 2 + 𝐶 𝑥 +1 𝑥 − 2 𝑥 + 1 𝑥 − 2 𝑥 + 2 By cancelling denominator 5𝑥 = 𝐴 𝑥−2 𝑥+2+𝐵 𝑥+1 𝑥+2+𝐶 𝑥+1 𝑥−2 Putting x = −1, in (1) 5 × −1 = 𝐴 −1−2 −1+2+𝐵 −1+1 −1+2+𝐶 −1+1 −1−2 −5 = 𝐴 −3 1+𝐵×0+𝐶×0 −5 = −3𝐴 𝐴 = − 5− 3 = 53 Similarly Putting x = 2, in (1) 5(2) = 𝐴 2−2 2+2+𝐵 2+1 2+2+𝐶 2+1 2−2 10 = 𝐴×0+𝐵 3 4+𝐶×0 10 = 12𝐵 𝐵 = 1012= 56 Similarly Putting x = −2 , in (1) 5(−2) = 𝐴 −2−2 −2+2+𝐵 −2+1 −2+2+𝐶 −2+1 −2−2 −10 = 𝐴×0+𝐵×0+𝐶 −1 −4 −10 = 4𝐶 𝐶 = −104 𝐶 = −52 Therefore 5𝑥 𝑥 + 1 𝑥2− 4= 𝐴𝑥 + 1+ 𝐵𝑥 − 2+ 𝐶𝑥 + 2𝑑𝑥 = 53 𝑑𝑥𝑥 + 1 𝑑𝑥+ 56 𝑑𝑥𝑥 − 2 𝑑𝑥− 52 𝑑𝑥 𝑥 + 2 = 𝟓𝟑 𝒍𝒐𝒈 𝒙+𝟏− 𝟓𝟐 𝐥𝐨𝐠 𝒙+𝟐+ 𝟓𝟔 𝐥𝐨𝐠 𝒙−𝟐+𝑪

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About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .