Ex 7.5, 14 - Chapter 7 Class 12 Integrals

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Ex 7.5, 14 3𝑥 − 1﷮ 𝑥 + 2﷯﷮2﷯﷯ We can write it as 3𝑥 − 1﷮ 𝑥 + 2﷯﷮2﷯﷯ = 𝐴﷮ 𝑥 + 2﷯﷯+ 𝐵﷮ 𝑥 + 2﷯﷮2﷯﷯ 3𝑥 − 1﷮ 𝑥 + 2﷯﷮2﷯﷯ = 𝐴 𝑥 + 2﷯ + 𝐵﷮ 𝑥 + 2﷯﷮2﷯﷯ 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 − 2﷯﷮2﷯﷯ = 3﷮𝑥 − 2﷯+ −7﷮ 𝑥 − 2﷯﷮2﷯﷯ Integrating 𝑤.𝑟.𝑡.𝑥. ﷮﷮ 3𝑥 − 1﷮ 𝑥 − 2﷯﷮2﷯﷯﷯ 𝑑𝑥= ﷮﷮ 3﷮𝑥 − 2﷯ − 5﷮ 𝑥 − 2﷯﷮2﷯﷯﷯﷯𝑑𝑥 =3 ﷮﷮ 1﷮𝑥 − 2﷯ ﷯𝑑𝑥−7 ﷮﷮ 1﷮ 𝑥 − 2﷯﷮2﷯﷯ ﷯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