# Ex 7.5, 14 - Chapter 7 Class 12 Integrals

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 7.5, 14 3𝑥 − 1 𝑥 + 22 We can write it as 3𝑥 − 1 𝑥 + 22 = 𝐴 𝑥 + 2+ 𝐵 𝑥 + 22 3𝑥 − 1 𝑥 + 22 = 𝐴 𝑥 + 2 + 𝐵 𝑥 + 22 By cancelling denominator 3𝑥−1 = 𝐴 𝑥 + 2+𝐵 Putting x = 2 in (1) 3𝑥−1=𝐴 𝑥+2+𝐵 3 −2−1=𝐴 −2+2+𝐵 −6−1=𝐴×0+𝐵 −7=𝐵 𝐵=−7 Putting x = 2 in (1) 3𝑥−1=𝐴 𝑥+2+𝐵 3×0−1=𝐴 0+2+𝐵 −1=2𝐴+𝐵 −1=2𝐴+ −7 −1+7=2𝐴 2𝐴=6 𝐴=3 Hence we can write it as 3𝑥 − 1 𝑥 − 22 = 3𝑥 − 2+ −7 𝑥 − 22 Integrating 𝑤.𝑟.𝑡.𝑥. 3𝑥 − 1 𝑥 − 22 𝑑𝑥= 3𝑥 − 2 − 5 𝑥 − 22𝑑𝑥 =3 1𝑥 − 2 𝑑𝑥−7 1 𝑥 − 22 𝑑𝑥 =3 log 𝑥−2−7 𝑥−2−2 𝑑𝑥 =3 log 𝑥−2−7 𝑥 − 2−2 + 1−2 + 1 =3 log 𝑥−2−7 𝑥 − 2−1−1 +𝐶 =𝟑 𝐥𝐨𝐠 𝒙−𝟐+ 𝟕 𝒙 − 𝟐 +𝑪

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.