Ex 7.5, 19 - Class 12

Transcript

Ex 7.5, 19 2𝑥﷮( 𝑥﷮2﷯ + 1)( 𝑥﷮2﷯ + 3)﷯ Let 𝑥﷮2﷯=𝑡 Differentiate both sides 𝑤.𝑟.𝑡.𝑥. 2𝑥 = 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥 = 𝑑𝑡﷮2𝑥﷯ Substituting value of 𝑡 & 𝑑𝑥 ﷮﷮ 2𝑥﷮( 𝑥﷮2﷯ + 1)( 𝑥﷮2﷯ + 3)﷯﷯ 𝑑𝑥 = ﷮﷮ 2𝑥 𝑑𝑡﷮ 𝑡 + 1﷯ 𝑡 + 3﷯﷯﷯ = ﷮﷮ 2𝑥 𝑑𝑡﷮ 𝑡 + 1﷯ 𝑡 + 3﷯2𝑥﷯﷯ = ﷮﷮ 𝑑𝑡﷮ 𝑡 + 1﷯ 𝑡 + 3﷯﷯﷯ Now we can write 1﷮ 𝑡 + 1﷯ 𝑡 + 3﷯﷯ = 𝐴﷮𝑡 + 1﷯ + 𝐵﷮𝑡 + 3﷯ 1﷮ 𝑡 + 1﷯ 𝑡 + 3﷯﷯ = 𝐴 𝑡 + 3﷯ + 𝐵 𝑡 + 1﷯﷮ 𝑡 + 1﷯ 𝑡 + 3﷯﷯ Cancelling denominator 1 = 𝐴 𝑡+3﷯ + 𝐵 𝑡+1﷯ Putting t = −1 in (1) 1 = 𝐴 −1+3﷯ + 𝐵 −1+1﷯ 1 = 𝐴×2+ 𝐵×0 1 = 2𝐴 𝐴 = 1﷮2﷯ Similarly Putting t=−3 in (1) 1 = 𝐴 𝑡+3﷯+ B (𝑡+1) 1 = 𝐴 −3+3﷯ + 𝐵 −3+1﷯ 1 = 𝐴×0+ 𝐵× −2﷯ 1 = −2𝐵 𝐵 = −1﷮2﷯ Therefore ﷮﷮ 1﷮ 𝑡 + 1﷯ 𝑡 + 3﷯﷯﷯ 𝑑𝑡 = ﷮﷮ 1﷮2 𝑡 + 1﷯﷯﷯ 𝑑𝑥 + ﷮﷮ −1﷮2 𝑡 + 3﷯﷯﷯ 𝑑𝑥 = 1﷮2﷯ log ﷮ 𝑡+1﷯﷯− 1﷮2﷯ log ﷮ 𝑡+3﷯﷯+𝐶 = 1﷮2﷯ log [﷮ 𝑡+1﷯﷯−log |t+3|]+C = 1﷮2﷯ log ﷮ 𝑡 + 1﷮𝑡 + 3﷯﷯﷯+𝐶 Substituting back the value of t = 𝑥﷮2﷯ = 1﷮2﷯ log ﷮ 𝑥﷮2﷯ + 1﷮ 𝑥﷮2﷯ + 3﷯﷯﷯+𝐶 = 𝟏﷮𝟐﷯ 𝒍𝒐𝒈 ﷮ 𝒙﷮𝟐﷯ + 𝟏﷮ 𝒙﷮𝟐﷯ + 𝟑﷯﷯﷯+𝑪