Ex 7.4, 1 - Chapter 7 Class 12 Integrals

Last updated at Dec. 16, 2024 by Teachoo

Ex 7.4, 1 - Integrate 3x2 / x6 + 1 - Chapter 7 Class 12 - Ex 7.4 - Ex 7.4

part 2 - Ex 7.4, 1 - Ex 7.4 - Serial order wise - Chapter 7 Class 12 Integrals

part 3 - Ex 7.4, 1 - Ex 7.4 - Serial order wise - Chapter 7 Class 12 Integrals

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Transcript

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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