Ex 7.4, 1 - Chapter 7 Class 12 Integrals

Last updated at Dec. 16, 2024 by Teachoo

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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