Slide7.JPG

Slide8.JPG
Slide9.JPG

This video is only available for Teachoo black users

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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 )

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.