Ex 7.5, 8 - Integrate x / (x - 1)2 (x + 2) - NCERT

Ex 7.5, 8 𝑥﷮ 𝑥 − 1﷯﷮2﷯ (𝑥 + 2)﷯ We can write the integrand as 𝑥﷮ 𝑥 − 1﷯﷮2﷯ (𝑥 + 2)﷯ = 𝐴﷮ 𝑥 − 1﷯﷯ + 𝐵﷮ 𝑥 − 1﷯﷮2﷯﷯ + 𝐶﷮ 𝑥 + 2﷯﷯ 𝑥﷮ 𝑥 − 1﷯﷮2﷯ (𝑥 + 2)﷯ = 𝐴 𝑥 − 1﷯ 𝑥 + 2﷯ + 𝐵 𝑥 + 2﷯ + 𝐶 𝑥 −1﷯﷮2﷯﷮ 𝑥 − 1﷯ 𝑥 + 1﷯ 2𝑥 + 3﷯﷯ By cancelling denominator 𝑥 = 𝐴 𝑥−1﷯ 𝑥+2﷯ + 𝐵 𝑥+2﷯ + 𝐶 𝑥−1﷯﷮2﷯ Putting x = −2, in (1) −2 = A −2−1﷯ −2+2﷯ + B −2+2﷯ + C −2−1﷯﷮2﷯ −2 = A×0+B×0+C −3﷯﷮2﷯ −2 = 9C C = − 2﷮9﷯ Putting x = 1, in (1) 1 = A 1−1﷯ 1+2﷯ + B 1+2﷯ + C 1−1﷯﷮2﷯ 1 = A×0 + B×3 + C 1−1﷯﷮2﷯ 1 = 3B B = 1﷮3﷯ Putting x = 0, in (1) 0=𝐴 0−1﷯ 0+2﷯+𝐵 0+2﷯ + 𝐶 0−1﷯﷮2﷯ 0=𝐴 −1﷯ 2﷯+2𝐵+𝐶 0=−2𝐴+ 2﷮3﷯ − 2﷮9﷯ 0=−2𝐴+ 4﷮9﷯ 2𝐴= 4﷮9﷯ ⇒𝐴= 4﷮18﷯ ⇒𝐴= 2﷮9﷯ Hence we can write ﷮﷮ 𝑥﷮ 𝑥 − 1﷯﷮2﷯ (𝑥 + 2)﷯﷯ 𝑑𝑥 = ﷮﷮ 2﷮9﷯﷮ 𝑥 − 1﷯﷯﷯ 𝑑𝑥 + ﷮﷮ 1﷮3﷯﷮ 𝑥 − 1﷯﷮2﷯﷯﷯ 𝑑𝑥 + ﷮﷮ − 2﷮9﷯﷮ 𝑥 + 2﷯﷯﷯ 𝑑𝑥 = 2﷮9﷯ ﷮﷮ 𝑑𝑥﷮ 𝑥 − 1﷯﷯﷯+ 1﷮3﷯ ﷮﷮ 𝑑𝑥﷮ 𝑥 − 1﷯﷮2﷯﷯﷯+ − 2﷮9﷯ ﷮﷮ 𝑑𝑥﷮ 𝑥 + 2﷯﷯﷯ = 2﷮9﷯ ﷮﷮ 𝑑𝑥﷮ 𝑥 − 1﷯﷯﷯ + 1﷮3﷯ ﷮﷮ 𝑑𝑥﷮ 𝑥 − 1﷯﷮2﷯﷯﷯ + − 2﷮9﷯ ﷮﷮ 𝑑𝑥﷮ 𝑥 + 2﷯﷯﷯ = 2﷮9﷯ log ﷮ 𝑥−1﷯﷯+ 1﷮3﷯ −1﷮ 𝑥 − 1﷯﷯﷯− 2﷮9﷯ log ﷮ 𝑥+2﷯﷯+𝐶 = 2﷮9﷯ log ﷮ 𝑥−1﷯﷯− 2﷮9﷯ log ﷮ 𝑥+2﷯﷯− 1﷮3 𝑥 − 1﷯﷯ +𝐶 = 2﷮9﷯ log ﷮ 𝑥−1﷯﷯− log ﷮ 𝑥+2﷯﷯﷯− 1﷮3 𝑥 − 1﷯﷯ +𝐶 = 𝟐﷮𝟗﷯ 𝐥𝐨𝐠 ﷮ 𝒙 − 𝟏﷮𝒙 + 𝟐﷯﷯﷯− 𝟏﷮𝟑 𝒙 − 𝟏﷯﷯ +𝑪