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Ex 7.4, 1 - Chapter 7 Class 12 Integrals (Term 2)

Last updated at Dec. 11, 2021 by

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Ex 7.4, 1 (3𝑥^2)/(𝑥^6 + 1) We need to find ∫1▒(𝟑𝒙^𝟐)/(𝒙^𝟔 + 𝟏) 𝒅𝒙 Let 𝒙^𝟑=𝒕 Diff both sides w.r.t. x 3𝑥^2=𝑑𝑡/𝑑𝑥 𝒅𝒙=𝒅𝒕/(𝟑𝒙^𝟐 ) Thus, our equation becomes ∫1▒(𝟑𝒙^𝟐)/(𝒙^𝟔 + 𝟏) 𝒅𝒙 =∫1▒(3𝑥^2)/((𝑥^3 )^2 + 1) 𝑑𝑥 Putting the value of 𝑥^3=𝑡 and 𝑑𝑥=𝑑𝑡/(3𝑥^2 ) =∫1▒(3𝑥^2)/(𝑡^2 + 1) .𝑑𝑡/(3𝑥^2 ) =∫1▒𝑑𝑡/(𝑡^2 + 1) =∫1▒𝒅𝒕/(𝒕^𝟐 + (𝟏)^𝟐 ) =1/1 tan^(−1)⁡〖 𝑡/1 〗+𝐶 It is of form ∫1▒𝑑𝑡/(𝑥^2 + 𝑎^2 ) =1/𝑎 〖〖𝑡𝑎𝑛〗^(−1) 〗⁡〖𝑥/𝑎〗 +𝐶 ∴ Replacing 𝑥 = 𝑡 and 𝑎 by 1 , we get =tan^(−1)⁡〖 (𝑡)〗+𝐶 =〖〖𝒕𝒂𝒏〗^(−𝟏) 〗⁡(𝒙^𝟑 )+𝑪 ("Using" 𝑡=𝑥^3 )

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.