Ex 7.5, 23 - Chapter 7 Class 12 Integrals

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Ex 7.5, 23 ﷮﷮ 𝑑𝑥﷮𝑥(𝑥2 + 1) ﷯﷯ equals • log 𝑥﷯ - 1﷮2 ﷯ log (x2+1) + C • log 𝑥﷯ + 1﷮2 ﷯ log (x2+1) + C • log 𝑥﷯ + 1﷮2 ﷯ log (x2+1) + C • 1﷮2 ﷯ log 𝑥﷯ + log (x2+1) + C By Multiplying integrand by 𝑥﷮𝑥﷯ ﷮﷮ 1﷮𝑥(𝑥2 + 1)﷯﷯ . 𝑥﷮𝑥﷯ 𝑑𝑥 = ﷮﷮ 𝑥 𝑑𝑥﷮ 𝑥﷮2﷯ 𝑥﷮2﷯ + 1﷯﷯﷯ Let t = 𝑥﷮2﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥 2𝑥 = 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥 = 𝑑𝑡﷮2𝑥﷯ Thus, our equation become ﷮﷮ 𝑥﷮ 𝑥﷮2﷯( 𝑥﷮2﷯ + 1)﷯﷯ 𝑑𝑥 = ﷮﷮ 𝑥﷮𝑡(𝑡 + 1)﷯﷯ 𝑑𝑡﷮2𝑥﷯ = ﷮﷮ 1﷮2﷯ 𝑑𝑡﷮𝑡(𝑡 + 1)﷯﷯ = 1﷮2﷯ ﷮﷮ 𝑑𝑡﷮𝑡(𝑡 + 1)﷯﷯ We can write integrand as 1﷮𝑡(𝑡 + 1)﷯ = 𝐴﷮𝑡﷯ + 𝐵﷮𝑡 + 1﷯ 1﷮𝑡(𝑡 + 1)﷯ = 𝐴(𝑡 + 1) + 𝐵𝑡﷮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 x = −1 in (1) 1 = 𝐴(−1+1)+𝐵 −1﷯ 1 = 𝐴×0−𝐵 1 = −𝐵 𝐵 = −1 Therefore 1﷮2﷯ ﷮﷮ 1﷮𝑡 𝑡 + 1﷯﷯﷯ 𝑑𝑡 = 1﷮2﷯ ﷮﷮ 1﷮𝑡﷯ ﷯𝑑𝑡 + 1﷮2﷯ ﷮﷮ −1﷮𝑡 + 1﷯ ﷯𝑑𝑡 = 1﷮2﷯ log ﷮ 𝑡﷯﷯− 1﷮2﷯ log ﷮ 𝑡+1﷯﷯+𝐶 Putting back t = 𝑥﷮2﷯ = 1﷮2﷯ log ﷮ 𝑥﷮2﷯﷯﷯− 1﷮2﷯ log ﷮ 𝑥﷮2﷯+1﷯﷯+𝐶 = 1﷮2﷯×2 log ﷮ 𝑥﷯﷯− 1﷮2﷯ log ﷮ 𝑥﷮2﷯+1﷯﷯+𝐶 = log ﷮ 𝑥﷯﷯− 1﷮2﷯ log ﷮( 𝑥﷮2﷯+1)﷯+𝐶 ∴ The correct answer is A .