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Ex 7.5, 10 2𝑥 − 3﷮ 𝑥2 − 1﷯ 2𝑥 + 3﷯﷯ We can write the integrand as 2𝑥 − 3﷮ 𝑥2 − 1﷯ 2𝑥 + 3﷯﷯ = 2𝑥 − 3﷮ 𝑥 − 1﷯ 𝑥 + 1﷯ 2𝑥 + 3﷯﷯ = 𝐴﷮ 𝑥 + 1﷯﷯ + 𝐵﷮ 𝑥 − 1﷯﷯ + 𝐶﷮ 2𝑥 + 3﷯﷯ = 𝐴 𝑥 − 1﷯ 2𝑥 + 3﷯ + 𝐵 𝑥 + 1﷯ 2𝑥 + 3﷯ + 𝐶 𝑥 −1﷯ 𝑥 + 3﷯﷮ 𝑥 − 1﷯ 𝑥 + 1﷯ 2𝑥 + 3﷯﷯ By cancelling denominator 2𝑥−3 = 𝐴 𝑥−1﷯ 2𝑥+3﷯+𝐵 𝑥+1﷯ 2𝑥+3﷯+𝐶 𝑥−1﷯ 𝑥+1﷯ By substituting x = − 1, in (1) −2−3 = 𝐴 −1−1﷯ −2+3﷯+𝐵 −1+1﷯ −2+3﷯+𝐶 −1+1﷯ −1−1﷯ −5 = 𝐴 −2﷯ 1﷯+𝐵×0+𝐶×0 −5 = −2𝐴 𝐴 = 5﷮2﷯ Similarly Putting x = 1, in (1) 2−3 = 𝐴 1−1﷯ 2+3﷯+𝐵 1+1﷯ 2+3﷯+𝐶 1−1﷯ 1+1﷯ −1 = A 0﷯+𝐵 2﷯ 5﷯+𝐶 0﷯ −1 = 10𝐵 𝐵 = − 1﷮10﷯ Similarly Putting x = − 3﷮2﷯ , in (1) 2 −3﷮2﷯﷯−3 = 𝐴 −3﷮2﷯ −1﷯ 2 −3﷮2﷯﷯+3﷯+𝐵 −3﷮2﷯+1﷯ 2 −3﷮2﷯﷯+3﷯+𝐶 −3﷮2﷯ −1﷯ −3﷮2﷯+1﷯ −6 = 𝐴×0+𝐵×0+𝐶 − 1﷮2﷯﷯× − 5﷮2﷯﷯ −6 = 5﷮4﷯ 𝐶 𝐶 = − 24﷮5﷯ Therefore 2𝑥 − 3﷮ 𝑥 + 1﷯ 𝑥 − 1﷯ 2𝑥 + 3﷯﷯ = 5﷮2﷯﷮ 𝑥 + 1﷯﷯ + − 1﷮10﷯﷯﷮ 𝑥 − 1﷯﷯ + − 24﷮5﷯﷯ ﷮ 2𝑥 + 3﷯﷯ Hence we can write it as ﷮﷮ 2𝑥 − 3﷮ 𝑥 + 1﷯ 𝑥 − 1﷯ 2𝑥 + 3﷯﷯﷯ = ﷮﷮ 5﷮2 𝑥 + 1﷯﷯﷯ 𝑑𝑥 + ﷮﷮ −1﷮10 𝑥 − 1﷯﷯﷯ 𝑑𝑥 + ﷮﷮ −24﷮5 2𝑥 + 3﷯﷯﷯ 𝑑𝑥 = 5﷮2﷯ log ﷮ 𝑥+1﷯﷯− 1﷮10﷯ log ﷮ 𝑥−1﷯﷯− 24﷮5﷯ × 1﷮2﷯ log ﷮ 2𝑥+3﷯﷯+𝐶 = 𝟓﷮𝟐﷯ 𝒍𝒐𝒈 ﷮ 𝒙+𝟏﷯﷯− 𝟏﷮𝟏𝟎﷯ 𝐥𝐨𝐠 ﷮ 𝒙−𝟏﷯﷯− 𝟏𝟐﷮𝟓﷯ 𝐥𝐨𝐠 ﷮ 𝟐𝒙+𝟑﷯﷯+𝑪