Ex 7.5, 16 - Class 12

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Ex 7.5, 16 1﷮𝑥 𝑥﷮𝑛﷯+1﷯﷯ [Hint: multiply numerator and denominator by xn − 1 and put xn = t] We can write integrand as 1﷮𝑥 𝑥﷮𝑛﷯ + 1﷯﷯ = 1﷮𝑥 𝑥﷮𝑛﷯ + 1﷯﷯ × 𝑥﷮𝑛 − 1﷯﷮ 𝑥﷮𝑛 − 1﷯﷯ = 𝑥﷮𝑛﷯ ﷮𝑥 × 𝑥﷮𝑛 − 1﷯ 𝑥﷮𝑛﷯ + 1﷯﷯ = 𝑥﷮𝑛 − 1﷯ ﷮ 𝑥﷮𝑛﷯ 𝑥﷮𝑛﷯ + 1﷯﷯ Let t = 𝑥﷮𝑛﷯ Then 𝑑𝑡﷮𝑑𝑥﷯ = 𝑛 𝑥﷮𝑛 − 1﷯ Therefore ﷮﷮ 𝑥﷮𝑛 − 1﷯ ﷮ 𝑥﷮𝑛﷯ 𝑥﷮𝑛﷯ + 1﷯﷯﷯ 𝑑𝑥 = ﷮﷮ 𝑥﷮𝑛 − 1﷯ ﷮𝑡 𝑡 + 1﷯﷯﷯ . 𝑑𝑡﷮𝑛 𝑥﷮𝑛 − 1﷯﷯ = 1﷮n﷯ ﷮﷮ 𝑑𝑡 ﷮𝑡 𝑡 + 1﷯﷯﷯ We can write the integrand as 1﷮𝑡 𝑡 + 1﷯﷯ = 𝐴﷮𝑡﷯ + 𝐵﷮𝑡 + 1﷯ 1﷮𝑡 𝑡 + 1﷯﷯ = 𝐴 𝑡 + 1﷯ + 𝐵 𝑡﷮𝑡 𝑡 + 1﷯﷯ By cancelling denominator 1 = 𝐴 𝑡 + 1﷯ +𝐵𝑡 Putting t = 0 in (1) 1 = 𝐴 0+1﷯+𝐵×0 1 = 𝐴 𝐴 = 1 Similarly putting t = −1 in (1) 1 = 𝐴 −1+1﷯+𝐵 −1﷯ 1 = 𝐴 0﷯−𝐵 −1 = 𝐵 𝐵 = −1 Therefore we can write 1﷮𝑡 𝑡 + 1﷯﷯ = 1﷮𝑡﷯− 1﷮𝑡 + 1﷯ Thus, ﷮﷮ 𝑑𝑡 ﷮𝑡 𝑡 + 1﷯﷯﷯ = ﷮﷮ 1 𝑑𝑡﷮𝑡﷯﷯ − ﷮﷮ −1﷮𝑡 + 1﷯﷯ 𝑑𝑡 = log ﷮ 𝑡﷯﷯− log ﷮ 𝑡+1﷯﷯ + C = log ﷮ 𝑡﷮𝑡 + 1﷯﷯﷯+𝐶 Putting back t = 𝑥﷮𝑛﷯ ﷮﷮ 1﷮𝑥 𝑥﷮𝑛﷯ + 1﷯﷯﷯ = 𝟏﷮𝐧﷯ 𝐥𝐨𝐠 ﷮ 𝒙﷮𝒏﷯﷮ 𝒙﷮𝒏﷯ + 𝟏﷯﷯﷯+𝑪